Statistics II Solved Paper 2012 FBISE

Statistics II Solved Paper 2012 FBISE
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In this article, titled “Statistics II Solved Paper 2012 FBISE,” we will walk through a comprehensive analysis of the solved paper, addressing various aspects of statistics as encountered in the 2013 FBISE exam. The post will cover multiple-choice questions (MCQs), short-answer questions, numerical problems, and extensive questions on fundamental topics such as probability distribution, binomial and hypergeometric distributions, normal distribution, sampling distribution, inferential statistics (hypothesis testing and estimation), and association using the Chi-square test.

Each topic will be explained in detail, with solutions to guide students through the process of answering such questions efficiently. Whether you’re preparing for an exam or just brushing up on your statistics knowledge, this post serves as a helpful resource for mastering key statistical concepts and methodologies. This topic is equally important for the students of statistics across all the major Boards and Universities such as FBISE, BISERWP, BISELHR, MU, DU, PU, NCERT, CBSE & others & across all the statistics, business & finance disciplines.

Statistics II Solved Paper 2012 FBISE

MCQS

Fill the relevant bubble against each question according to curriculum:
1In how many ways a team of four players can be chosen from 1 o players?
A)40B)210
C)5040D)None of these
2The events A and 8 are mutually exclusive then
A)A U B = SB)A ∩ B = A
C)A ∩ B = ϕD)None of these
3If the random variable “X” denotes the number of heads when three distinct coins are tossed, then “X” assumes the value
A)1,2,3B)1,3,3, 1
C)0, 1,2,3D)None of these
4If E(X) = 1.5, Var(X) = 0.5 then E(X²) will be:
A)2.75B)2.25
C)0.25D)2.7
5In a multiple choice test there are five possible answers to each of 20 questions If a candidate guesses the correct answer each time, what is the mean number of the correct answer?
A)5B)1/5
C)4D)20
6Given N = 11 n= 5 K = 6, then P ( x ≥ 1) equals:
A)1B)1/66
C)65/66D)461/462
7Standard deviation of standard normal distribution is:
A)1B)0
C)1.2D)None of these
8In normal probability distribution P(z < -1) = 0.1587 then P (z>1) is:
A)0.6587B)0.1587
C)0.8413D)0.3413
9The measure on sample is called:
A)StatisticsB)Population
C)StatisticD)Parameter
10The standard deviation of sampling distribution is called:
A)Sampling errorB)Non-sampling error
C)Standard errorD)Bias
11A population contains two Items and four items are selected at random with replacement, then all possible samples shall be
A)16B)8
C)

    \[ \begin{pmatrix}4 \\C \\2 \\\end{pmatrix}\ \]

 		D)4
12A statistic ɵ̂ is an estimator of parameter ɵ, then it will be unbiased if:
A)E(ɵ̂)>ɵB)E(ɵ̂)=ɵ
C)E(ɵ̂)<ɵD)E(ɵ̂)≠ɵ
13For large sample size (n>30) the value of Z α/2  for 90% confidence interval shall be:
A)±1.96B)±1.645
C)±2.33D)±2.58
14In testing hypothesis about difference between population means for small samples and unknown standard deviations, the degree of freedom shall be:
A)n-1B)n1+n2-2 
C)n-2 D)2n-2 
15Ho is true and we reject it is called:
A)Type II errorB)Standard Error
C)Type I errorD)Sampling Error
16For α= 0.0 I the critical value for Z = ——- for one tailed test are equal to:
A)– 1.96 or 1.96B)– 2.33 or 2.33
C)– 1.645 or 1.645D)None of these
17A binary digit is commonly called:
A)BitB)Byte
C)KilobyteD)Gigabyte
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Short Questions

(i) Distinguish between Permutation & Combination.

Answer:

Permutation

In permutation, order of the elements matters. For example, if there is a specific number 256, which means strictly 256, neither 526 nor 652. Permutation is used when order of the items is important. In mathematics permutation can be calculated through following formula:

    \[ \begin{pmatrix}\mathbf{n} \\\mathbf{P} \\\mathbf{r} \\\end{pmatrix}\mathbf{=}\frac{\mathbf{n!}}{\left( \mathbf{n - r}\right)\mathbf{!}}\ \]

Combination

In combination, order of the elements does not matter. For example, if there is one basket containing apples, bananas and oranges, no matter what comes first either banana, apples, oranges or oranges, bananas and apples etc. Combination is used when order of the elements is not important. Formula for combination is:

    \[ \begin{pmatrix}\mathbf{n} \\\mathbf{C} \\\mathbf{r} \\\end{pmatrix}\mathbf{=}\frac{\mathbf{n!}}{\mathbf{r!}\left( \mathbf{n- r} \right)\mathbf{!}}\ \]

(ii) What is the sample space when three distinct coins are tossed? Also specify the event A for exactly two heads.

Solution

Sample Space η(S) = 2³= 8

All possible samples =

HHHHHTHTHTHH
TTTTTHTHTHTT

Event: Exactly 2 Heads η(A) = 3

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\ \]

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{3}}{\mathbf{8}}\ \]

    \[ \mathbf{P}\left( \mathbf{A} \right)\) \textbf{= 0.375} \ \]

(iii) What is Probability density function? What are its properties?

Answer:

Probability density function (PDF) is a function that describes the likelihood of continuous random variable on a particular value. PDF determines probabilities in interval rather than a single point.

Properties of Probability Density Function

  • Non Negative

The probability of an outcome cannot be negative.

    \[ \mathbf{f}\left( \mathbf{x} \right)\mathbf{\ \geq 0\ for\ all\ x\ values}\ \]

  • Total Probability equals to one

Total area of PDF is equals to one in which we assume that random variable will take some value.

    \[ \int_{\mathbf{- \infty}}^{\mathbf{\infty}}{\mathbf{f}\left( \mathbf{x} \right)\mathbf{dx = 1\ }}\ \]

  • Probability of single value is zero

In PDF, the probability of a single value is zero.

    \[ \mathbf{P(X = x) = 0}\ \]

  • PDF defines the probabilities in interval

As we discussed above that total area of PDF is equals to one and probabilities are calculated in intervals.

    \[ \mathbf{P}\left( \mathbf{a \leq X \leq b} \right)\mathbf{=}\int_{\mathbf{a}}^{\mathbf{b}}{\mathbf{f}\left( \mathbf{x} \right)\mathbf{dx}}\) \ \]

(iv) Write a set A containing all vowels in the word “PUBLICATION”. Then find probability of A. Also find the probability of consonants.

Solution:

Set of Vowels (U, I, A, I, O) Set of Consonants (P, B, L, C, T, N)

Sample Space η(S) =

Event: Vowels η(A) = 5

Event: Consonants η(B) = 6

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\ \]

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{5}}{\mathbf{11}}\ \]

P(A)=0.455 

    \[ \mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\ \]

    \[ \mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{6}}{\mathbf{11}}\ \]

P(B) = 0.5455

    \[ \mathbf{P}\left( \mathbf{A} \right)\) \textbf{= 0.5455} \(\mathbf{or\ P}\left( \mathbf{B} \right)\mathbf{= \ 1 - P}\left( \mathbf{A} \right)\ \]

    \[ \mathbf{\ 1 - 0.4545 = 0.5455}\ \]

(v) Given a random variable “X” with E(X) = 2.5 and Variance(X) = 1.25 then find E(2x + 3), E(x²) & Var(2x – 5)

Solution

E(X) = 2.5, Variance(X) = 1.25

  • E(2x + 3) = 2E(X) + 3 = 2 (2.5) + 3 = 5+3 = 8
  • Variance (X) = E(X)² – [E(X)]²

1.25 = E(X)² – (2.5)²

E(X)² = 1.25 + 6.25

E(X)² = 1.25 + 6.25

E(X)² = 7.5

  • Var(2x – 5) = E(2X – 5)² – [E(2X – 5)]²

Where E(2X – 5) = 2E(X) – 5 = 2(2.5) – 5 = 5 – 5 = 0 &

E(2X – 5)² = E(4X² – 20X + 25) = 4(7.5) – 20(2.5) + 25 = 30 – 50 + 25 = 5

Var(2x – 5) = 5 – (0)² = 5

(vi) What is Binomial distribution & what are its properties?

Answer: Binomial distribution is a function which is used to calculate the binomial experiment.

Function is given below:

    \[ \mathbf{P(X\ = \ x)\ = \ }\mathbf{b(x,\ n,\ p)\ }\begin{pmatrix}\mathbf{n} \\\mathbf{x} \\\end{pmatrix}\mathbf{\ }\mathbf{p}^{\mathbf{x}}\mathbf{\ }\mathbf{q}^{\mathbf{n - x}}\ \]

for x = 0, 1, 2, 3 …..n.

Where n = sample size, x = number of successes, n – x = number of failure

P =success and q = failure = 1 – p

Properties

  1. The mean of binomial distribution µ = np
  2. Variance 𝜎² = npq
  3. Standard Deviation σ= √npq
  4. ∑px = 1
  5. If p = q = ½ then probability distribution is symmetrical
  6. For negative skewness P>q
  7. For positive skewness P<q

(vii) If X is a binomial random variable with mean 2.4 & variance 0.96, find P(X = 0)

Solution

Mean = np = 2.4 & Var = npq = 0.96

npq = npq

2.4q = 0.96       

q = 0.96/2.4 = 0.4

If q = 0.4 then p = 1 – q = 1 – 0.4 = 0.6

npq = npq

n(0.6)(0.4) = 0.96

n = 0.96/0.24 = 4

So now n = 4, p = 0.6, q = 0.4 and x = 0

    \[ \mathbf{P(X\ = \ x)\ = \ b(x,\ n,\ p)}\begin{pmatrix}\mathbf{n} \\\mathbf{x} \\\end{pmatrix}\mathbf{\ }\mathbf{p}^{\mathbf{x}}\mathbf{\ }\mathbf{q}^{\mathbf{n - x}}\ \]

    \[ \mathbf{P(X\ = \ 0)\ = \ b(0,\ 4,\ 0.6)\ }\begin{pmatrix} \mathbf{4} \\ \mathbf{0} \\ \end{pmatrix}\mathbf{\ }\mathbf{(0.6)}^{\mathbf{0}}\mathbf{\ }\mathbf{(0.4)}^{\mathbf{4 - 0}}\ \]

    \[ \mathbf{P(X\ = \ 0)\ = \ 0.0256}\ \]

(viii) In normal distribution lower & upper quartiles are 15 & 25, respectively. Find mean & standard deviation of the distribution. Also find mean deviation.

Solution

As we know that:

Q1 = µ – 0.6745 𝜎

Q3 = µ + 0.6745 𝜎

15 =  µ – 0.6745 𝜎

25 = µ + 0.6745 𝜎 sigma cancels each other

40 = 2µ

µ = 40/2 = 20

Now put the value of µ in any above equation to get 𝜎

15 = µ – 0.6745 𝜎

15 = 20 – 0.6745 𝜎

20 – 0.6745 𝜎 = 15

– 0.6745 𝜎 = 15 – 20

– 0.6745 𝜎 = – 5

𝜎 = 5/0.6745 = 7.4129

(ix) What is Standardized normal variate? Write down the equation of standardized normal distribution. Find the value of maximum ordinate.

Answer: The standardized Normal Variate or SNV which converts or transforms a normal random variable in to standard normal form. It is denoted by Z and it follows the standard normal distribution having mean zero (0) and standard deviation (1). Its formula is given below:

    \[Z = \ \frac{X - \ \mu}{\sigma}\]

Where:

Z = Standardized normal variate

X = Observed value

μ = Mean of the distribution

σ = Standard deviation of the distribution

Equation of the Normal Curve for SNV:

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The equation for the standard normal curve (probability density function) is:

    \[\mathbf{f}\left( \mathbf{Z} \right)\mathbf{= \ }\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi}}}\mathbf{\ }\mathbf{e}^{\mathbf{-}\frac{\mathbf{z}^{\mathbf{2}}}{\mathbf{2}}}\]

Where:

Z = Standardized normal variate

e = Euler’s number (approximately 2.71828)

π = Mathematical constant (approximately 3.14159)

Maximum Ordinate

Peak value of the pdf of standard normal distribution is at Z = 0 is called maximum ordinate:

    \[ \mathbf{f}\left( \mathbf{Z} \right)\mathbf{= \ }\frac{\mathbf{1}}{\sqrt{\mathbf{2}\mathbf{\pi}}}\mathbf{\ }\mathbf{e}^{\mathbf{-}\frac{\mathbf{z}^{\mathbf{2}}}{\mathbf{2}}}\ \]

    \[ \mathbf{f}\left( \mathbf{0} \right)\mathbf{= \ }\frac{\mathbf{1}}{\sqrt{\mathbf{2(3.14159}}}\mathbf{\ }\mathbf{2.71828}^{\mathbf{-}\frac{\mathbf{0}^{\mathbf{2}}}{\mathbf{2}}}\ \]

    \[ \mathbf{f}\left( \mathbf{0} \right)\mathbf{= 0.3989}\ \]

(x) Define a Random Sample. How would you obtain it?

Answer:

Random samples are sample points chosen randomly from population. Each sample point has equal chance to be selected. it must be ensured in random sampling that each sample should be chosen independently, has equal chance to be selected, each sample should not influence the other and each sample should represent the population.

Random Samples can be obtained through the following methods:

1. Simple Random Sampling

2. Systematic Random Sampling

3. Stratified Random Sampling

4. Cluster Random Sampling

5. Random Number Table or Software

(xi) If n = 5, N = 25 & 𝜎x̅ = 10 then find 𝜎²

Solution

If 𝜎x̅ = 10 then 𝜎²x̅ = 100

    \[ \mathbf{\sigma}^{\mathbf{2}}\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sigma ²}}{\mathbf{n}}\mathbf{\ }\left\lbrack \frac{\mathbf{N - n}}{\mathbf{N - 1}} \right\rbrack\ \]

    \[ \mathbf{100\ =}\frac{\mathbf{\sigma ²}}{\mathbf{5}}\mathbf{\ }\left\lbrack \frac{\mathbf{25 - 5}}{\mathbf{25 - 1}} \right\rbrack\ \]

    \[ \mathbf{100\ =}\frac{\mathbf{\sigma ²}}{\mathbf{5}}\mathbf{\ }\left\lbrack \mathbf{0.84} \right\rbrack\ \]

    \[ \left\lbrack \frac{\mathbf{100}}{\mathbf{0.84}} \right\rbrack\mathbf{= \ }\frac{\mathbf{\sigma ²}}{\mathbf{5}}\ \]

    \[ \mathbf{119\ \times 5\ = \ }\mathbf{\sigma}\mathbf{²}\ \]

    \[ \mathbf{\sigma}\mathbf{²\ = \ 595\ Approx}\ \]

(xii) Differentiate between Point estimation & Interval estimation.

Answer:

POINT ESTIMATION

A point estimate of a population parameter is a single value of a statistic. For example,
the sample mean x̅ is a point estimate of the population mean μ. Similarly, the sample proportion p is a point estimate of the population proportion P.

INTERVAL ESTIMATION

An interval estimate is defined by two numbers, between which a population parameter is said to lie.For example, a < x < b is an interval estimate of the population mean μ. It indicates that the population mean is greater than a bit less than b.

(xiii) What is an Unbiased estimator?

Answer:

As we know that sample is a small representative part of population. Similarly, sample statistic is a representative of population parameter. So when sample statistic such as sample mean is equal to the population mean without any bias, it is called unbiased estimator otherwise it is called biased estimator.

(xiv) Given (AB) = 150 (αB) = 106 (Aβ) = 272 (αβ) = 1132 & N = 1660. Find the coefficient of association.

Solution

    \[ \mathbf{Q =}\frac{\left( \mathbf{AB} \right)\left( \mathbf{\alpha\beta} \right)\mathbf{- \ (A\beta)(\alpha B)}}{\left( \mathbf{AB} \right)\left( \mathbf{\alpha\beta} \right)\mathbf{+ \ (A\beta)(\alpha B)}}\ \]

    \[ \mathbf{Q = \ }\frac{\left( \mathbf{150} \right)\left( \mathbf{1132} \right)\mathbf{- \ (272)(106)}}{\left( \mathbf{150} \right)\left( \mathbf{1132} \right)\mathbf{+ \ (272)(106)}}\ \]

    \[ \mathbf{Q = \ }\frac{\mathbf{169800 - \ 28832}}{\mathbf{169800 + \ 28832}}\mathbf{\ }\ \]

    \[ \mathbf{Q =}\frac{\mathbf{140968}}{\mathbf{198632}}\mathbf{= 0.709\ Positive\ Association}\ \]

(xv) Determine whether the two attributes A & B are independent, positively associated or negatively associated (A) = 490 (α) = 570 (AB) = 294 (αB) = 380

Solution

 AαTotal
B(AB) = (294)(αB) = (380)(B) = 674
β(A β) = (196)(αβ) = (190)(β) = 386
Total(A) = (490)(α) = (570)N or n = 1060

Here:

    \[AB\ = \ 294\]

    \[\frac{(A)(B)}{n} = \frac{(490)(674)}{1060} = 311.56\]

Conclusion

    \[ AB < \frac{(A)(B)}{n}\ \]

    \[ 294 < 311.56\ \]

Negative Association between A’ s and B’s

(xvi) Define one tailed & two tailed test.

Answer:

One-Tailed or One Sided Test

It is also called directional test. This type of test is used to test the significance and determine if there is a relationship between the variables in one direction.

Two-Tailed or Two Sided Test

It is also called non-directional test. This type of test is used to test the significance and determine if there is a relationship between the variables in either direction. In two tailed test α is divided by two and taking half on each side of the bell curve.

(xvii) Describe the procedure for testing hypothesis about mean of a normal population when population standard deviation is known.

Answer:

Sometimes the hypothesis about the population which is normal and its standard deviation 𝜎 is known. In this case Z test is used both for small and large sample size. Thus Z test is given below:

    \[ \mathbf{Z =}\frac{\overline{\mathbf{x}}\mathbf{- \mu}}{\mathbf{\sigma/}\sqrt{\mathbf{n}}}\ \]

where X̅ = Sample mean, µ = Population mean, 𝜎 = population standard deviation and n = sample size

(xviii) Given X̅1 = 26, X̅2 = 18, 𝜎x̅1 – x̅2 = 3.41,  Ho: µ1≤µ2 & α = 0.05. Find “Z” and make statistical decision.

Solution

Step 1 Hypothesis

Ho: µ1≤µ2

H1: µ1>µ2

Step 2 Level of Significance

α = 0.05

Step 3 Test Statistic

    \[ \mathbf{Z =}\frac{\left( {\overline{\mathbf{x}}}_{\mathbf{1}}\mathbf{-}{\overline{\mathbf{x}}}_{\mathbf{2}} \right)\mathbf{- \ (}\mathbf{\mu}_{\mathbf{1}}\mathbf{- \ }\mathbf{\mu}_{\mathbf{2}}\mathbf{)}}{\mathbf{\sigma}_{\mathbf{x\overline{}1\ - \ x\overline{}2}}}\) \ \]

Step 4 Critical Region

    \[ Tabulated\ Value\ Z_{\alpha} = Z_{0.05} = 1.645\ \]

    \[ where\ 50 - 5 = 0.45\ \]

Critical Value Z > 1.645

Step 5 Calculation of Z Statistic

    \[ \mathbf{Z =}\frac{\left( {\overline{\mathbf{x}}}_{\mathbf{1}}\mathbf{-}{\overline{\mathbf{x}}}_{\mathbf{2}} \right)\mathbf{- \ (}\mathbf{\mu}_{\mathbf{1}}\mathbf{- \ }\mathbf{\mu}_{\mathbf{2}}\mathbf{)}}{\mathbf{\sigma}_{\mathbf{x\overline{}1\ - \ x\overline{}2}}}\ \]

    \[ \mathbf{Z =}\frac{\left( \mathbf{26 - 18} \right)\mathbf{- \ (0)}}{\mathbf{3.41}}\mathbf{= 2.35}\ \]

Step 6 Conclusion

Calculated Z value falls in rejection region or greater than tabulated value so we reject Ho.

(xix) In a binomial distribution with n = 5 P(x = 0) = P(x = 1). Find the variance.

Solution

Here n = 5 and P(X=0) = P(X=1). Therefore

    \[ \mathbf{P}\left( \mathbf{X\ = 0} \right)\mathbf{= \ }\begin{pmatrix}\mathbf{5} \\\mathbf{0} \\\end{pmatrix}\mathbf{\ }\mathbf{p}^{\mathbf{0}}\mathbf{\ }\mathbf{q}^{\mathbf{5 - 0}}\mathbf{=}\mathbf{q}^{\mathbf{5}}\ \]

    \[ \mathbf{P}\left( \mathbf{X\ = 1} \right)\mathbf{= \ }\begin{pmatrix} \mathbf{5} \\\mathbf{1} \\\end{pmatrix}\mathbf{\ }\mathbf{p}^{\mathbf{1}}\mathbf{\ }\mathbf{q}^{\mathbf{5 -1}}\mathbf{=}{{\mathbf{5}\mathbf{p}}^{\mathbf{1}}\mathbf{q}}^{\mathbf{4}}\ \]

If P(x = 0) = P(x = 1) then:

    \[ \mathbf{q}^{\mathbf{5}}\mathbf{=}{{\mathbf{5}\mathbf{p}}^{\mathbf{1}}\mathbf{q}}^{\mathbf{4}}\ \]

    \[ \frac{\mathbf{q}^{\mathbf{5}}}{\mathbf{q}^{\mathbf{4}}}\mathbf{= 5}\mathbf{p = q = 5}\mathbf{p}\ \]

    \[ \mathbf{As\ we\ know\ that\ q = 1 - p\ so\ 1 - p = 5}\mathbf{p}\ \]

    \[ \mathbf{1 = 5}\mathbf{p + p = 1 = 6}\mathbf{p\ or\ p = \ }\frac{\mathbf{1}}{\mathbf{6}}\ \]

    \[ \mathbf{if\ p =}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{\ then\ q = \ }\frac{\mathbf{5}}{\mathbf{6}}\mathbf{\ }\ \]

    \[ \mathbf{Var}\left( \mathbf{X} \right)\mathbf{= \ npq}\ \]

    \[ \mathbf{Var}\left( \mathbf{X} \right)\mathbf{= \ 5}\left( \frac{\mathbf{1}}{\mathbf{6}} \right)\left( \frac{\mathbf{5}}{\mathbf{6}} \right)\mathbf{=}\frac{\mathbf{25}}{\mathbf{36}}\ \]

Extensive Questions

Q.3 (a) The probability is 2/3 that Mr. A will pass the examination and the probability is ¾ that Mr. B will pass the examination. Find the following probabilities: (i) Both will pass the examination, (ii) Only one will pass the examination, (iii) Somebody will pass the examination

Solution:

Data:

P(A will Pass the Examination) = P(A) =2/3, P(A will not Pass the Examination) =P(A̅)=1/3

P(B will Pass the Examination) = P(B) =3/4, P(B will not Pass the Examination) =P(B̅)=1/4

    \[ \left( \mathbf{i} \right)\mathbf{\ P}\left( \mathbf{Both\ will\ pass\ the\ Exam} \right)\mathbf{=}\mathbf{P}\left( \mathbf{A \cap B} \right)\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap B} \right)\mathbf{= P}\left( \mathbf{A} \right)\mathbf{P}\left( \mathbf{B} \right)\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap B} \right)\mathbf{=}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)\left( \frac{\mathbf{3}}{\mathbf{4}} \right)\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap B} \right)\mathbf{=}\frac{\mathbf{6}}{\mathbf{12}}\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap B} \right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{= 0.5}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ }\mathbf{P(Only\ One\ will\ Pass)\ =}\mathbf{P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{+ P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{+ P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\mathbf{=}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)\left( \frac{\mathbf{1}}{\mathbf{4}} \right)\mathbf{+}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)\left( \frac{\mathbf{3}}{\mathbf{4}} \right)\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{+ P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\mathbf{=}\frac{\mathbf{2}}{\mathbf{12}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{12}}\mathbf{=}\frac{\mathbf{5}}{\mathbf{12}}\ \]

    \[ \mathbf{(iii)\ 2/3\ \ that\ A\ will\ Pass\ and\ ¾\ that\ B\ will\ Pass}\ \]

(b) A continuous random variable “X” that can assume values between X = 2 and X = 5 has a density function given by f(x) = k (x + 1)

Solution:

    \[ \int_{\mathbf{a}}^{\mathbf{b}}{\mathbf{f}\left( \mathbf{x} \right)\mathbf{dx = 1}}\ \]

Where a = 2, and b = 5

    \[ \mathbf{f}\left( \mathbf{x} \right)\mathbf{= k}\left( \mathbf{x + 1} \right)\mathbf{\ where\ 2 \leq x \leq 5}\ \]

    \[ \int_{\mathbf{2}}^{\mathbf{5}}{\mathbf{f}\left( \mathbf{x} \right)\mathbf{dx = 1}}\ \]

Substitute f(x) = k(x+1)

    \[ \int_{\mathbf{2}}^{\mathbf{5}}{\mathbf{k}\left( \mathbf{x + 1} \right)\mathbf{dx = 1}}\ \]

    \[ \int_{\mathbf{2}}^{\mathbf{5}}{\mathbf{k}\left( \mathbf{x + 1} \right)\mathbf{dx = k}}\int_{\mathbf{2}}^{\mathbf{5}}{\left( \mathbf{x + 1} \right)\mathbf{dx}}\ \]

    \[ \int_{\mathbf{2}}^{\mathbf{5}}{\left( \mathbf{x + 1} \right)\mathbf{dx =}}\int_{\mathbf{2}}^{\mathbf{5}}{\mathbf{xdx\ + \ }\int_{\mathbf{2}}^{\mathbf{5}}{\mathbf{1}\mathbf{dx}}}\ \]

    \[ \int_{\mathbf{2}}^{\mathbf{5}}{\mathbf{xdx =}\left\lbrack \frac{\mathbf{x}^{\mathbf{2}}}{\mathbf{2}} \right\rbrack_{\mathbf{2}}^{\mathbf{5}}\mathbf{=}\frac{\mathbf{5}^{\mathbf{2}}}{\mathbf{2}}\mathbf{-}\frac{\mathbf{2}^{\mathbf{2}}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{25}}{\mathbf{2}}\mathbf{-}\frac{\mathbf{4}}{\mathbf{2}}\mathbf{=}\frac{\mathbf{21}}{\mathbf{2}}}\ \]

    \[ \int_{\mathbf{2}}^{\mathbf{5}}{\mathbf{1}\mathbf{dx =}\left\lbrack \mathbf{x} \right\rbrack_{\mathbf{2}}^{\mathbf{5}}\mathbf{= 5 - 2 = 3}}\ \]

    \[ \int_{\mathbf{2}}^{\mathbf{5}}{\left( \mathbf{x + 1} \right)\mathbf{dx =}\frac{\mathbf{21}}{\mathbf{2}}\mathbf{+ 3 =}\frac{\mathbf{27}}{\mathbf{2}}}\mathbf{\ }\ \]

Solving for k

    \[ \mathbf{k \times}\frac{\mathbf{27}}{\mathbf{2}}\mathbf{= 1}\ \]

    \[ \mathbf{k =}\frac{\mathbf{2}}{\mathbf{27}}\ \]

The density function is:

    \[ \mathbf{f}\left( \mathbf{x} \right)\mathbf{=}\frac{\mathbf{2}}{\mathbf{27}}\left( \mathbf{x + 1} \right)\mathbf{,\ 2 \leq x \leq 5}\ \]

(c) Four coins are tossed. A tail as recorded as 2 and head as 1 The variable of interest is the product of records. Find the probability distribution of this random variable Also find its mean and variance.

Solution:

All Possible Outcomes

H,H,H,HH,H,H,TH,H,T,HH,T,H,HT,H,H,H
T,T,T,TT,T,T,HT,T,H,TT,H,T,TH,T,T,T
H,T,H,TT,H,T,HH,H,T,TT,T,H,HH,T,T,H
T,H,H,T    

Tail recorded as 2 and Head as 1

1,1,1,11,1,1,21,1,2,11,2,1,12,1,1,1
2,2,2,22,2,2,12,2,1,22,1,2,21,2,2,2
1,2,1,22,1,2,11,1,2,22,2,1,11,2,2,1
2,1,1,2    

Product Table

12222
168888
44444
4    
Xp(x)xp(x)x²p(x)
11/161/161/16
24/168/161616
46/1624/1696/16
84/1632/16256/16
161/1616/16256/16
  ∑ xp(x) = 81/16∑ x²p(x) = 625/16

Mean (E(X) = ∑ xp(x) = 81/16

Mean (E(X) = 5.0625

Var(x) = ∑ x²p(x) – [∑ xp(x)]²

Var(x) = 625/16 – (5.0625)²

Var(x) = 13.4336

Q.4 (a) Five dice are thrown together 243 times Find the expected frequencies when throwing a three or four’s regarded as success. Calculate the mean and variance of this distribution of the number of three’s and four’s.

Solution:

Data

The number of dice thrown at a time: n=5

The number of trials: N=243

Success is defined as rolling a 3 or 4 on a single die.

Probability of success (rolling 3 or 4)

    \[ \mathbf{p =}\frac{\mathbf{2}}{\mathbf{6}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\ \]

    \[ \mathbf{q = 1 - p = 1 -}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}\ \]

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Binomial Distribution

    \[ \mathbf{P}\left( \mathbf{X = x} \right)\mathbf{= N}\begin{pmatrix}\mathbf{n} \\\mathbf{x}\\\end{pmatrix}\mathbf{p}^{\mathbf{x}}\mathbf{q}^{\mathbf{n - x}}\ \]

    \[ \mathbf{P}\left( \mathbf{X = 0} \right)\mathbf{= 243}\begin{pmatrix}\mathbf{5} \\\mathbf{0} \\\end{pmatrix}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)^{\mathbf{0}}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)^{\mathbf{5 - 0}}\ \]

    \[ \mathbf{P}\left( \mathbf{X = 0} \right)\mathbf{= 243}\left( \mathbf{1} \right)\left( \mathbf{1} \right)\left( \frac{\mathbf{32}}{\mathbf{243}} \right)\mathbf{= 32}\ \]

    \[ \mathbf{P}\left( \mathbf{X = 1} \right)\mathbf{= 243}\begin{pmatrix}\mathbf{5} \\\mathbf{1} \\\end{pmatrix}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)^{\mathbf{1}}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)^{\mathbf{5 - 1}}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{1} \right)\mathbf{= 243}\left( \mathbf{5} \right)\left( \frac{\mathbf{1}}{\mathbf{3}} \right)\left(\frac{\mathbf{16}}{\mathbf{81}} \right)\mathbf{=}\mathbf{80}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{2} \right)\mathbf{= 243}\begin{pmatrix}\mathbf{5} \\\mathbf{2} \\\end{pmatrix}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)^{\mathbf{2}}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)^{\mathbf{5 -}\mathbf{2}}\ \]

    \[ \mathbf{P}\left( \mathbf{X = 2} \right)\mathbf{= 243}\left( \mathbf{10} \right)\left( \frac{\mathbf{1}}{\mathbf{9}} \right)\left(\frac{\mathbf{8}}{\mathbf{27}} \right)\mathbf{= 80}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{3} \right)\mathbf{= 243}\begin{pmatrix}\mathbf{5} \\\mathbf{3} \\\end{pmatrix}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)^{\mathbf{3}}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)^{\mathbf{5 -}\mathbf{3}}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{3} \right)\mathbf{= 243}\left( \mathbf{10} \right)\left( \frac{\mathbf{1}}{\mathbf{27}} \right)\left( \frac{\mathbf{4}}{\mathbf{9}} \right)\mathbf{=}\mathbf{4}\mathbf{0}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{4} \right)\mathbf{= 243}\begin{pmatrix}\mathbf{5} \\\mathbf{4} \\\end{pmatrix}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)^{\mathbf{4}}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)^{\mathbf{5 -}\mathbf{4}}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{4} \right)\mathbf{= 243}\left( \mathbf{5} \right)\left( \frac{\mathbf{1}}{\mathbf{81}} \right)\left( \frac{\mathbf{2}}{\mathbf{3}} \right)\mathbf{=}\mathbf{1}\mathbf{0}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{5} \right)\mathbf{= 243}\begin{pmatrix}\mathbf{5} \\\mathbf{5} \\\end{pmatrix}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)^{\mathbf{5}}\left( \frac{\mathbf{2}}{\mathbf{3}} \right)^{\mathbf{5 -}\mathbf{5}}\ \]

    \[ \mathbf{P}\left( \mathbf{X =}\mathbf{5} \right)\mathbf{= 243}\left( \mathbf{1} \right)\left( \frac{\mathbf{1}}{\mathbf{243}} \right)\left( \mathbf{1} \right)\mathbf{= 1}\ \]

Frequency Distribution

XFP(x)
03232/243
18080/243
28080/243
34040/243
41010/243
511/243
 ∑f=243∑p(x) = 243/243 = 1

    \[ \mathbf{Mean\ = \ np\ = \ }\left( \mathbf{5} \right)\left( \frac{\mathbf{1}}{\mathbf{3}} \right)\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}}\ \]

    \[ \mathbf{V}\mathbf{ar(x)}\mathbf{\ = \ npq\ }\ \]

    \[ \mathbf{Var}\left( \mathbf{x} \right)\mathbf{= 5}\left( \frac{\mathbf{1}}{\mathbf{3}} \right)\left( \frac{\mathbf{2}}{\mathbf{3}} \right)\mathbf{=}\frac{\mathbf{10}}{\mathbf{9}}\ \]

b. A machine dispenses liquid into bottles in such a way that amount of liquid dispensed on each occasion is normally distributed with mean and standard deviation 266 and 20ml respectively Bottles that weigh less than 260 ml have to be recycled. What percentage of bottles should be recycled?

Solution:

Step 1: Data

  • Mean (μ): 266 ml
  • Standard deviation (σ): 20 ml
  • Threshold for recycling: X=260 ml

Step 2: Test Statistic

The z-score is calculated as:

    \[ \mathbf{z =}\frac{\mathbf{X - \mu}}{\mathbf{\sigma}}\ \]

Step 3: Calculation

    \[ \mathbf{z} = \frac{\mathbf{X} - \mathbf{\mu}}{\mathbf{\sigma}}\ \]

    \[ \mathbf{z =}\frac{\mathbf{260}\mathbf{-}\mathbf{266}}{\mathbf{20}}\mathbf{= - 0.3}\ \]

Results:

Z-score: −0.3

Percentage of bottles to be recycled: 38.21%

Conclusion:

The z-score of −0.3 corresponds to a cumulative probability of approximately 38.21%. This means that 38.21% of the bottles are dispensed with less than 260 ml of liquid and will need to be recycled. ​

Q.5 (a) If mean and variance of population are 7 and 3.15. What would be standard error. If samples are drawn without replacement of size 6. If number of population unit is 10.

Solution:

Data

    \[ \mathbf{\mu = 7,\ }\mathbf{\sigma}^{\mathbf{2}}\mathbf{= 3.15,}\mathbf{\ \sigma = 1.7748,}\mathbf{\ n = 6,N = 10}\ \]

    \[ \mathbf{S.E = \ }\mathbf{\sigma}_{\overline{\mathbf{x}}}\ \]

Verification Formula

    \[ \mathbf{\sigma}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{\sigma}}{\sqrt{\mathbf{n}}}\sqrt{\frac{\mathbf{N - n}}{\mathbf{N - 1}}}\ \]

    \[ \mathbf{\sigma}_{\overline{\mathbf{x}}}\mathbf{=}\frac{\mathbf{1.7748}}{\sqrt{\mathbf{6}}}\sqrt{\frac{\mathbf{10 - 6}}{\mathbf{10}\mathbf{- 1}}}\ \]

    \[ {\mathbf{S.E =}\mathbf{\sigma}}_{\overline{\mathbf{x}}}\mathbf{=}\mathbf{0.483}\ \]

b. A random sample of 200 workers was selected from a population and 140 workers were found to be skilled. The factory owner claimed that at least 80% workers were skilled in his factory. Is It possible to reject the claim of factory owner at 5% level of significance?

Solution:

Step 1: Hypothesis

Null Hypothesis Ho: p ≥ 0.80

Alternative Hypothesis H1: p < 0.80

Step 2: Level of Significance

α = 0.05

Step 3: Test Statistic

    \[ \mathbf{Z =}\frac{\widehat{\mathbf{p}}\mathbf{- po}}{\sqrt{\left( \frac{\mathbf{po\ qo}}{\mathbf{n}} \right)}}\ \]

Step 4: Critical Region

    \[ Z \textless{} - 1.645 \]

    \[ \mathbf{Z}_{\mathbf{\alpha}}\mathbf{=}\mathbf{Z}_{\mathbf{0.05}}\mathbf{= - 1.645)}\ \]

Step 5: Computations

Here n=200, X=140 (Skilled workers)

    \[ \mathbf{p}_{\mathbf{0}}\mathbf{= 0.}\mathbf{8}\mathbf{0\ }\ \]

    \[ \mathbf{q}_{\mathbf{0}}\mathbf{= 1 -}\mathbf{p}_{\mathbf{0}}\ \]

    \[ \mathbf{q}_{\mathbf{0}}\mathbf{=}\mathbf{1 - 0.}\mathbf{8}\mathbf{0}\ \]

    \[ \mathbf{q}_{\mathbf{0}}\mathbf{= 0.}\mathbf{2}\mathbf{0}\ \]

    \[ \widehat{\mathbf{p}}\mathbf{=}\frac{\mathbf{14}\mathbf{0}}{\mathbf{2}\mathbf{00}}\mathbf{= 0.7}\ \]

    \[ \mathbf{Z =}\frac{\widehat{\mathbf{p}}\mathbf{- po}}{\sqrt{\left( \frac{\mathbf{po\ qo}}{\mathbf{n}} \right)}}\ \]

    \[ \mathbf{Z =}\frac{\mathbf{0.7 - 0.}\mathbf{8}\mathbf{0}}{\sqrt{\left( \frac{\left( \mathbf{0.}\mathbf{8}\mathbf{0} \right)\left( \mathbf{0.}\mathbf{2}\mathbf{0} \right)}{\mathbf{2}\mathbf{00}} \right)}}\ \]

    \[ \mathbf{Z}\mathbf{=}\frac{\mathbf{- 0.}\mathbf{1}}{\mathbf{0.0}\mathbf{282}}\ \]

    \[ \mathbf{Z}\mathbf{= -}\mathbf{3.546}\ \]

Step 6: Conclusion

Our calculated value of Z = -3.546 falls in rejection or critical region so we will reject Null Hypothesis and accept Alternative Hypothesis. We can conclude that skilled workers are less than 80%.

c. A random sample of 200 married men was classified according to education and number of children. Test the hypothesis at 5% level of significance that the size of family as independent of the level of education, attained by fathers.

EducationNumber of Children
0-12-3Over 3
Elementary143732
Secondary194317
College221710

Solution:

EducationNumber of ChildrenTotal
0-10-10-1
Elementary14373283
Secondary19431779
College22171049
Total559759211

Step 1 Testing the Hypothesis

Ho: There is no Association between Level of Education & Number of Children.

H1: There is Association between Level of Education & Number of Children.

Step 2 Level of Significance

Level of Significance = α=0.05

Step 3 Test Statistic

    \[ \mathbf{\chi ² = \ \sum}\frac{\mathbf{(fo - fe)}^{\mathbf{2}}}{\mathbf{fe}}\ \]

Step 4 Critical Region

Critical Region: Degree of Freedom d.f= (R-1)(C-1)

So d.f= (3-1)(3-1)=4

The Value of Tabulated χ²(0.01,4)=13.277

The Critical Region χ²cal>13.277

Step 5 Calculation of Expected Frequencies

    \[ \frac{83 \times 55}{211} = 22,\ \frac{83 \times 97}{211} = 38,\ \frac{83 \times 59}{211} = 23\ \]

    \[ \frac{79 \times 55}{211} = 21,\ \frac{79 \times 97}{211} = 36,\ \frac{79 \times 59}{211} = 22\ \ \]

    \[ \frac{49 \times 55}{211} = 13,\ \frac{49 \times 97}{211} = 23,\ \frac{49 \times 59}{211} = 13\ \ \]

Step 6 Calculation of χ²

Table B. Computation of χ²
fofefo-fe(fo-fe)²

    \[ \frac{\left( \mathbf{fo - fe} \right)^{\mathbf{2}}}{\mathbf{fe}}\ \]

1422-8642.909
3738-110.026
32239813.522
1921-240.190
43367491.361
1722-5251.136
22139816.231
1723-6361.565
1013-390.692
    17.633
    

    \[ \mathbf{\sum}\left\lbrack \frac{\left( \mathbf{fo - fe} \right)^{\mathbf{2}}}{\mathbf{fe}} \right\rbrack\mathbf{=}\ \]

Step 7 Conclusion

The calculated value of χ² is 17.633 is more than the tabulated value of χ² 13.277 or 17.633 falls in the rejection region. We reject the Null Hypothesis Ho, accepts alternative hypothesis H1 and conclude that there is association between Level of Education & Number of Children

Statistics II Solved Paper 2012 FBISE

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Statistics II Solved Paper 2013 FBISE

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