Statistics II Solved Paper 2007 FBISE

Statistics II Solved Paper 2007 FBISE

In this article, titled “Statistics II Solved Paper 2007 FBISE,” we will walk through a comprehensive analysis of the solved paper, addressing various aspects of statistics as encountered in the 2007 FBISE exam. The post will cover multiple-choice questions (MCQs), short-answer questions, numerical problems, and extensive questions on fundamental topics such as probability distributionbinomial and hypergeometric distributionsnormal distributionsampling distributioninferential 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 FBISEBISERWPBISELHRMUDUPU, NCERT, CBSE & others & across all the statistics, business & finance disciplines.

Table of Contents

Statistics II Solved Paper 2007 FBISE

MCQS

Fill the relevant bubble against each question according to curriculum:
1A and B are independent events if P(A/B) = P(B)
A)TrueB)False
2P(A) = ¾ means that odds are 3 to 4 that event A will occur:
A)TrueB)False
3Probability of drawing a red card from a pack of 52 cards is 26/52:
A)TrueB)False
4For two independent variables X and Y Var(X – Y) = Var(X) – Var(Y):
A)TrueB)False
5E(6X + 15) = 6E(X)+…..
A)TrueB)False
6A binomial distribution will be symmetrical if p and q are:
A)DifferentB)Distinct
C)IdenticalD)None of these
7For a binomial distribution ∑(q+p) = 1
A)TrueB)False
8Area under standard normal curve is…..
A)1B)0
C)100D)10
9Normal distribution and Gaussian distribution are two different distributions:
A)TrueB)False
10The factor is called the finite population correction factor.
A)n/NB)

    \[ \sqrt{\frac{\mathbf{N - n}}{\mathbf{N - 1}}}\ \]

C)N/nD)None of these
11Sampling error is the difference between the population parameter and the
A)parameterB)sample statistic
C)sampleD)None of these
12For α = 0.01 the critical value of z = …. For two tailed test.
A)2.58B)1.96
C)2.33D)5.96
13If a deserving player is not selected in the team it is….
A)Type I ErrorB)Type II Error
C)Correct DecisionD)None of these
14If (AB) = [(A)(B)]/N then A and B are said to be independent.
A)TrueB)False
15Ho: µ ≥ 30 is a …… hypothesis.
A)One sidedB)Two sided
C)AlternativeD)None of these
Statistics II Solved Paper 2007 FBISE
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Short Questions

(i) Explain Exhaustive Events & Independent Events.

Answer:

Exhaustive Events

The events are said to be exhaustive when they are such that at least one of the events compulsorily occurs. For example, in throwing of coin tail or head must be occurred.

Independent Events

The events are said to be independent when they have no relation with each other or we can say that if event A has no impact on event B to happen then they are independent event.

(ii) Consider two events A & B such that P(A̅) = 1/3, P(B/A) = 3/5 and P(B/A̅) = 4/5. Find P(B).

Solution

As we know that P(A/B) = P(A) & P(B/A) = P(B)

    \[ \textbf{And} \mathbf{P(A \cap B) = P(B \cap A)}\ \]

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

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

    \[ \mathbf{P(B \cap A) = \ }\frac{\mathbf{3}}{\mathbf{5}}\mathbf{\  \times \ }\frac{\mathbf{2}}{\mathbf{3}}\mathbf{\  = \ }\frac{\mathbf{6}}{\mathbf{15}}\ \]

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

    \[ \frac{\mathbf{2}}{\mathbf{3}}\mathbf{= \ }\frac{\frac{\mathbf{6}}{\mathbf{15}}}{\mathbf{P}\left( \mathbf{B} \right)}\ \]

    \[ \mathbf{P}\left( \mathbf{B} \right)\left( \frac{\mathbf{2}}{\mathbf{3}} \right)\mathbf{= \ }\frac{\mathbf{6}}{\mathbf{15}}\ \]

    \[ \mathbf{P}\left( \mathbf{B} \right)\left( \frac{\mathbf{2}}{\mathbf{3}} \right)\mathbf{= \ }\frac{\frac{\mathbf{6}}{\mathbf{15}}}{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{= \ }\frac{\mathbf{3}}{\mathbf{5}}\ \]

    \[ \mathbf{Hence\ \ proved\ that\ P}\left( \frac{\mathbf{B}}{\mathbf{A}} \right)\mathbf{= \ P(B)}\ \]

(iii) What is density function? Write down 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) If a variable has the binomial distribution  determine its Variance.

Solution

    \[ \mathbf{Variance = npq}\ \]

where:

n: number of trials

p: probability of success

q: probability of failure (q=1−p)

(v) Give four important properties of normal distribution.

Answer:

  1. The mean, median and mode are equal.
  2. Mean deviation is 0.7979 of its standard deviation.
  3. Quartile Deviation is 0.6745 of its standard deviation.
  4. The curve is symmetrical about ordinate at X̅ = µ.
  5. The two quartiles are at equal distance from the mean and are at distance of 0.6745 𝜎.

(vi) Define & Explain the following (a) Parameter & Statistic (b) Sample Survey

Answer:

(a) Parameter & Statistic

Statistic is a statistical measure of sample, for example X̅ and S are statistical measure of sample whereas parameter is a statistical measure of population for example µ and 𝜎 are parameters.

(b) Sample Survey

Sample survey is a data gathering survey which is conducted to obtain data in order to calculate sample statistic to know or estimate the population parameter. For example, sample mean is an estimator of population mean which can be calculated through sample survey.

(vii) What is meant by estimator & estimate?

Answer:

Estimator

An estimator is a sample statistic which estimates some facts about the population. You can also think of an estimator as the rule that creates an estimate. For example, the sample mean (x̄) is an estimator for the population mean, μ. Similarly, sample variance S² is an estimate of population variance σ².

Estimate

Estimate with the help of estimator is called estimate. There are two types of estimates (i) point estimate (ii) interval estimate.

Point Estimate

A point estimate of a population parameter is a single value used to estimate the population parameter. For example, the sample mean x̅ is a point estimate of the population mean μ.

Interval Estimate

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

(viii) Define (a) Level of Significance and Critical Region (b) One tailed & Two Tailed Tests.

Answer:

Level of Significance & Critical Region

The probability of rejecting true null hypothesis is called level of significance and it is denoted by α, Level of significance is also called Type I error.

The critical region is the range of values for a test statistic where the null hypothesis (Ho​) is rejected. It is determined based on the level of significance (α) and corresponds to the extreme regions of the probability distribution. If the test statistic falls within the critical region, the result is considered statistically significant.

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.

(ix) Explain the difference between (a) Attributes & Variable (b) Positive & Negative Association.

Answer:

Attributes & Variable

Attribute is an observed qualitative property which can be expressed in qualitative aspect. Attribute has descriptive nature such as color, gender, quality etc. It may be discrete or categorical such as Yes, No or Pass, Fail.

Variable is qualitative measure which can be expressed in terms of quantity such as income, saving, consumption, profit, price etc. there are multiple types of variables dependent, independent, discrete or continuous etc.

Positive & Negative Association

If two variables move in same direction either positive or negative, association will be positive whereas if they move in opposite direction, the association will be negative.

(x) Two dice are rolled, find the probability of getting similar faces.

Solution

Sample Space η(S) = 6² = 36

Event: Similar Faces η(E) = 6

All possible outcomes

1, 12, 13, 14, 15, 16, 1
1, 22, 23, 24, 25, 26, 2
1, 32, 33, 34, 35, 36, 3
1, 42, 43, 44, 45, 46, 4
1, 52, 53, 54, 55, 56, 5
1, 62, 63, 64, 65, 66, 6

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

    \[ \mathbf{P}\left( \mathbf{E} \right)\mathbf{= \ }\frac{\mathbf{6}}{\mathbf{36}}\ \]

    \[ \mathbf{P}\left( \mathbf{E} \right)\mathbf{= \ }\frac{\mathbf{1}}{\mathbf{6}}\mathbf{= \ 0.167}\ \]

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Extensive Questions

Q.3 The probability that a man will be alive in 35 years is 3/5 and that his wife will be alive in 35 years is 2/3. Find the probability that: (i) Neither will be alive (ii) Both will be alive (iii) Only one will alive (iv) At least one will be alive (v) At most one will be alive in 35 years.

Solution

Probability of man will alive P(A) 3/5 then Probability of man will not alive P(A̅) 2/5

Probability of wife will alive P(B) 2/3 then Probability of wife will not alive P(B̅) 1/3

    \[ \left( \mathbf{i} \right)\mathbf{P}\left( \mathbf{Neither\ will\ be\ alive} \right)\mathbf{= \ P}\left( \overline{\mathbf{A}} \right)\mathbf{+ \ P}\left( \overline{\mathbf{B}} \right)\ \]

    \[ \mathbf{P(}\overline{\mathbf{A}}\mathbf{)\  + \ P(}\overline{\mathbf{B}}\mathbf{)\  =}\frac{\mathbf{2}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{11}}{\mathbf{15}}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{P}\left( \mathbf{Both\ will\ be\ alive} \right)\mathbf{= \ P}\left( \mathbf{A} \right)\mathbf{+ \ P}\left( \mathbf{B} \right)\ \]

    \[ \mathbf{P(A)\  + \ P(B) =}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{19}}{\mathbf{15}}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{P}\left( \mathbf{Only\ one\ will\ be\ alive} \right)\mathbf{= \ P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{\cup P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{\cup P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\mathbf{= \ }\left( \frac{\mathbf{3}}{\mathbf{5}}\mathbf{\times \ }\frac{\mathbf{1}}{\mathbf{3}} \right)\mathbf{+}\left( \frac{\mathbf{2}}{\mathbf{5}}\mathbf{\times \ }\frac{\mathbf{2}}{\mathbf{3}} \right)\ \]

    \[ \mathbf{P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{\cup P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\mathbf{= \ }\frac{\mathbf{3}}{\mathbf{15}}\mathbf{+ \ }\frac{\mathbf{4}}{\mathbf{15}}\mathbf{= \ }\frac{\mathbf{7}}{\mathbf{15}}\ \]

    \[ \left( \mathbf{iv} \right)\mathbf{P(At\ least\ one\ will\ be\ alive)\  = \ P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{\cup P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\mathbf{\cup P(A \cap B)}\ \]

    \[ \left( \frac{\mathbf{3}}{\mathbf{5}}\mathbf{\times \ }\frac{\mathbf{1}}{\mathbf{3}} \right)\mathbf{+}\left( \frac{\mathbf{2}}{\mathbf{5}}\mathbf{\times \ }\frac{\mathbf{2}}{\mathbf{3}} \right)\mathbf{+}\left( \frac{\mathbf{3}}{\mathbf{5}}\mathbf{\times \ }\frac{\mathbf{2}}{\mathbf{3}} \right)\ \]

    \[ \mathbf{\ }\frac{\mathbf{3}}{\mathbf{15}}\mathbf{+ \ }\frac{\mathbf{4}}{\mathbf{15}}\mathbf{+ \ }\frac{\mathbf{6}}{\mathbf{15}}\mathbf{= \ }\frac{\mathbf{13}}{\mathbf{15}}\ \]

    \[ \left( \mathbf{v} \right)\mathbf{P(At\ Most\ one\ will\ be\ alive)\  = \ P}\left( \mathbf{A \cap}\overline{\mathbf{B}} \right)\mathbf{\cup P}\left( \overline{\mathbf{A}}\mathbf{\cap B} \right)\mathbf{\cup P(}\overline{\mathbf{A}}\mathbf{\cap}\overline{\mathbf{B}}\mathbf{)}\ \]

    \[ \left( \frac{\mathbf{3}}{\mathbf{5}}\mathbf{\times \ }\frac{\mathbf{1}}{\mathbf{3}} \right)\mathbf{+}\left( \frac{\mathbf{2}}{\mathbf{5}}\mathbf{\times \ }\frac{\mathbf{2}}{\mathbf{3}} \right)\mathbf{+}\left( \frac{\mathbf{2}}{\mathbf{5}}\mathbf{\times}\frac{\mathbf{1}}{\mathbf{3}} \right)\mathbf{\ }\ \]

    \[ \frac{\mathbf{3}}{\mathbf{15}}\mathbf{+ \ }\frac{\mathbf{4}}{\mathbf{15}}\mathbf{+ \ }\frac{\mathbf{2}}{\mathbf{15}}\mathbf{= \ }\frac{\mathbf{9}}{\mathbf{15}}\ \]

Q.4 Five balls are drawn from a bag containing 4 white and 7 black balls. If X denotes the number of black balls drawn obtain the probability distribution of X. Find coefficient of variation of the distribution.

Solution

Data

White Balls = 4, Black Balls = 7, N = 4+7 = 11, n = 5, X = 0, 1, 2, 3, 4, 5

Sample Space

    \[ \mathbf{\eta\ }\left( \mathbf{S} \right)\mathbf{= \ }\begin{pmatrix}\mathbf{N} \\\mathbf{C} \\\mathbf{n} \\\end{pmatrix}\mathbf{= \ }\begin{pmatrix}\mathbf{11} \\\mathbf{C} \\\mathbf{5} \\\end{pmatrix}\mathbf{= 462}\ \]

Events

    \[ \mathbf{\ (X\  = \ 1)\  = \ }\begin{bmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{4} \\\end{bmatrix}\begin{bmatrix}\mathbf{7} \\\mathbf{C} \\\mathbf{1} \\\end{bmatrix}\mathbf{= \ 1\  \times \ 7 = 7}\ \]

    \[ \mathbf{P}\left( \mathbf{X\  = \ 1}\right)\mathbf{=}\frac{\mathbf{7}}{\mathbf{462}}\ \]

    \[ \mathbf{(X\  = \ 2)\  = \ }\begin{bmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{3} \\\end{bmatrix}\begin{bmatrix}\mathbf{7} \\\mathbf{C} \\\mathbf{2} \\\end{bmatrix}\mathbf{= \ 4\  \times \ 21 = 84}\ \]

    \[ \mathbf{P}\left( \mathbf{X\  = \ 2}\right)\mathbf{=}\frac{\mathbf{84}}{\mathbf{462}}\ \]

    \[ \mathbf{(X\  = \ 3)\  = \ }\begin{bmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{2} \\\end{bmatrix}\begin{bmatrix}\mathbf{7} \\\mathbf{C} \\\mathbf{3} \\\end{bmatrix}\mathbf{= \ 6\  \times \ 35 = 210}\ \]

    \[ \mathbf{P}\left( \mathbf{X\  = \ 3}\right)\mathbf{=}\frac{\mathbf{210}}{\mathbf{462}}\ \]

    \[ \mathbf{(X\  = \ 4)\  = \ }\begin{bmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{1} \\\end{bmatrix}\begin{bmatrix}\mathbf{7} \\\mathbf{C} \\\mathbf{4} \\\end{bmatrix}\mathbf{= \ 4\  \times \ 35 = 140}\ \]

    \[ \mathbf{P}\left( \mathbf{X\  = \ 4}\right)\mathbf{=}\frac{\mathbf{140}}{\mathbf{462}}\ \]

    \[ \mathbf{(X\  = \ 5)\  = \ }\begin{bmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{0} \\\end{bmatrix}\begin{bmatrix}\mathbf{7} \\\mathbf{C} \\\mathbf{5} \\\end{bmatrix}\mathbf{= \ 1\  \times \ 21 = 21}\ \]

    \[ \mathbf{P}\left( \mathbf{X\  = \ 5}\right)\mathbf{=}\frac{\mathbf{21}}{\mathbf{462}}\ \]

XP(X)xp(x)x²p(x)
17/4627/4627/462
284/462168/462336/462
3210/462630/4621890/462
4140/462560/4622240/462
521/462105/462525/462
  ∑xp(x) = 1470/462∑x²p(x) = 4998/462

    \[ \mathbf{E}\left( \mathbf{X} \right)\mathbf{= \ \sum xp}\left( \mathbf{x} \right)\mathbf{=}\frac{\mathbf{1470}}{\mathbf{462}}\mathbf{\  = \ 3.1818}\ \]

    \[ \mathbf{Var}\left( \mathbf{X} \right)\mathbf{= \ \sum}\mathbf{x}^{\mathbf{2}}\mathbf{p}\left( \mathbf{x} \right)\mathbf{-}\left\lbrack \mathbf{\sum xp}\left( \mathbf{x} \right) \right\rbrack^{\mathbf{2}}\mathbf{\ \ \ }\ \]

    \[ \mathbf{Var}\left( \mathbf{X} \right)\mathbf{= \ 10.818\ -\ }\left( \mathbf{3.1818} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{Var}\left( \mathbf{X} \right)\mathbf{= \ 10.818\ -\ 10.1238\ }\ \]

    \[ \mathbf{Var(X) = \ 0.6942}\ \]

    \[ \mathbf{S.D}\left( \mathbf{X} \right)\mathbf{= \ }\sqrt{\mathbf{Var}\left( \mathbf{X} \right)}\ \]

    \[ \mathbf{S.D}\left( \mathbf{X} \right)\mathbf{= \ }\sqrt{\mathbf{0.6942}}\ \]

    \[ \mathbf{S.D(X) = 0.8331\ \ }\ \]

    \[ \mathbf{C.V = \ }\left( \frac{\mathbf{S.D}}{\mathbf{Mean}} \right)\mathbf{100}\ \]

    \[ \mathbf{C.V = \ }\left( \frac{\mathbf{0.8331}}{\mathbf{3.1818}} \right)\mathbf{100}\ \]

    \[ \mathbf{C.V = \ 26.183\%\ }\ \]

Q.5 A certain event is believed to follow the binomial distribution. In 1024 samples of 5, the result was observed once in 405 times and twice in 270 times. Calculate the probability p and q.

Solution:

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

    \[ \mathbf{where\ x = 0,1,2,3,\ldots\ldots n.}\ \]

Here N = 1024, n = 5, x = 1 and 2

    \[ \mathbf{1024.P}\left\lbrack \mathbf{X = 1} \right\rbrack\mathbf{= 1024}\begin{pmatrix}\mathbf{5} \\\mathbf{1} \\\end{pmatrix}\mathbf{p\ }\mathbf{q}^{\mathbf{4}}\mathbf{= 405}\ \]

    \[ \mathbf{1024.P}\left\lbrack \mathbf{X = 2} \right\rbrack\mathbf{=1024}\begin{pmatrix}\mathbf{5} \\\mathbf{2} \\\end{pmatrix}\mathbf{p}^{\mathbf{2}}\mathbf{\ }\mathbf{q}^{\mathbf{3}}\mathbf{= 270}\ \]

    \[ \mathbf{5120\ p\ }\mathbf{q}^{\mathbf{4}}\mathbf{= 405\ldots..(i)}\ \]

    \[ \mathbf{10240\ p}^{\mathbf{2}}\mathbf{\ }\mathbf{q}^{\mathbf{3}}\mathbf{=270\ldots\ldots..(ii)}\ \]

Dividing equation ii by equation i

    \[ \frac{\mathbf{10240\ p}^{\mathbf{2}}\mathbf{\ }\mathbf{q}^{\mathbf{3}}}{\mathbf{5120\ p\ }\mathbf{q}^{\mathbf{4}}}\mathbf{=}\frac{\mathbf{270}}{\mathbf{405}}\ \]

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

    \[ \mathbf{2}\mathbf{p\  \times \ 3 = 2}\mathbf{q}\ \]

    \[ \mathbf{6}\mathbf{p = 2}\mathbf{q\ or\ 6}\mathbf{p = 2(1 - p)}\ \]

    \[ \mathbf{6}\mathbf{p + 2}\mathbf{p = 2}\ \]

    \[ \mathbf{8}\mathbf{p = 2}\ \]

    \[ \mathbf{p=}\frac{\mathbf{2}}{\mathbf{8}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{\ and\ q = 1 - p = 1 }\frac{\mathbf{1}}{\mathbf{4}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}\ \]

Q.6      In a normal distribution with µ = 47.6 and  = 16.2. Find. (i) The probability that a single observation will be larger than 55. (ii) Two points such that a single observation has a 90% probability of falling between them (iii) D7 (iv) P95

Solution (i)

    \[ \mathbf{Here\ µ\  = \ 47.6,\ \sigma\  = \ 16.2\ }\ \]

    \[ \mathbf{Z = \ }\frac{\mathbf{X - \ \mu}}{\mathbf{\sigma}}\mathbf{\  = \ }\frac{\mathbf{X - \ 47.6}}{\mathbf{16.2}}\ \]

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

    \[ \mathbf{Z = \ }\frac{\mathbf{55 - \ 47.6}}{\mathbf{16.2}}\mathbf{= 0.4567}\ \]

P(X>55) = P(Z>0.4567) = 0.5 + 1.712 = 2.212

(ii) 90% area under normal curve means that 0.45 area each to the left and right of mean. The value of z for 0.45 is 1.645. X1 will be left on the mean so Z1 will be negative and X2 will be left on the mean so Z2 will be positive.

Solution (ii)

    \[ \mathbf{Z}_{\mathbf{1}}\mathbf{= \ }\frac{\mathbf{X}\mathbf{1 - \ \mu}}{\mathbf{\sigma}}\ \]

    \[ \mathbf{- 1.645 = \ }\frac{\mathbf{X}_{\mathbf{1}}\mathbf{- \ 47.6}}{\mathbf{16.2}}\ \]

    \[ \mathbf{X}_{\mathbf{1}}\mathbf{\ -\ 47.6\  = \  - 1.645\  \times \ 16.2}\ \]

    \[ \mathbf{X}_{\mathbf{1}}\mathbf{\ -\ 47.6\  = \  - 26.649}\ \]

    \[ \mathbf{X}_{\mathbf{1}}\mathbf{\  = \  - 26.649\  + \ 47.6\  = \ 20.951}\ \]

    \[ \mathbf{Z}_{\mathbf{2}}\mathbf{= \ }\frac{\mathbf{X}\mathbf{1 - \ \mu}}{\mathbf{\sigma}}\ \]

    \[ \mathbf{1.645 = \ }\frac{\mathbf{X}_{\mathbf{2}}\mathbf{- \ 47.6}}{\mathbf{16.2}}\ \]

    \[ \mathbf{X}_{\mathbf{2}}\mathbf{\ -\ 47.6\  = \ 1.645\  \times \ 16.2}\ \]

    \[ \mathbf{X}_{\mathbf{2}}\mathbf{\ -\ 47.6\  = \ 26.649}\ \]

    \[ \mathbf{X}_{\mathbf{2}}\mathbf{\  = \ 26.649\  + \ 47.6\  = \ 74.249}\ \]

Solution (iii)

Here µ = 47.6,  = 16.2

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

    \[ \mathbf{Z = \ }\frac{\mathbf{X - \ 47.6}}{\mathbf{16.2}}\ \]

D7 = P70 is a point having 70% of the area below it. Area table shows this point to be:

0.7 – 0.5= 0.2 = 0.523

    \[ \mathbf{0.523 = \ }\frac{\mathbf{X - \ 47.6}}{\mathbf{16.2}}\ \]

    \[ \mathbf{X\ -\ 47.6\  = \ 0.523\  \times \ 16.2}\ \]

    \[ \mathbf{X\ -\ 47.6\  = \ 8.4726}\ \]

    \[ \mathbf{X\  = \ 8.4726\  + \ 47.6\  = \ 56.0726}\ \]

Solution (iv)

Here µ = 47.6,  = 16.2

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

    \[ \mathbf{Z = \ }\frac{\mathbf{X - \ 47.6}}{\mathbf{16.2}}\ \]

P95 is a point having 95% of the area below it. Area table shows this point to be:

0.95 – 0.5= 0.45 = 1.645

    \[ \mathbf{1.645 = \ }\frac{\mathbf{X - \ 47.6}}{\mathbf{16.2}}\ \]

    \[ \mathbf{X\ -\ 47.6\  = \ 1.645\  \times \ 16.2}\ \]

    \[ \mathbf{X\ -\ 47.6\  = \ 26.649}\ \]

    \[ \mathbf{X\  = \ 26.649\  + \ 47.6\  = \ 74.249}\ \]

Q.7A population consists of the number 2, 4, 8, 10, 12, 16 draw all possible samples of size 2 with replacement. Make a probability distribution of samples means and verify that:

    \[ \mathbf{(i)\ \mu}\overline{\mathbf{x}}\mathbf{= \ \mu\ }\ \]

    \[ \mathbf{(ii)\ }\mathbf{\sigma}^{\mathbf{2}}\overline{\mathbf{x}}\mathbf{= \ }\frac{\mathbf{\sigma}^{\mathbf{2}}}{\mathbf{n}}\mathbf{\ }\ \]

SolutionPopulation = 2, 4, 8, 10, 12, 16, Population Size N = 6, Sample size n = 2

    \[ \mathbf{Possible\ Samples\ = \ }\mathbf{N}^{\mathbf{n}}\mathbf{=}\mathbf{6}^{\mathbf{2}}\mathbf{= 36}\ \]

S/NoSamplesSum of SamplesMean of SamplesS/NoSamplesSum of SamplesMean of Samples
12, 2421910, 2126
22, 4632010, 4147
32, 81052110, 8189
42, 101262210, 102010
52, 121472310, 122211
62, 161892410, 162613
74, 2632512, 2147
84, 4842612, 4168
94, 81262712, 82010
104, 101472812, 102211
114, 121682912, 122412
124, 1620103012, 162814
138, 21053116, 2189
148, 41263216, 42010
158, 81683316, 82412
168, 101893416, 102613
178, 1220103516, 122814
188, 1624123616, 163216
Sampling Distribution of X̅
ffX̅fX̅²
2124
32618
41416
521050
6424144
7428196
8324192
9436324
10550500
11222242
12336432
13226338
14228392
16116256
Sum363123104
∑f =∑fX̅ =∑fX̅²=

Mean, Variance & S.D of Sampling Distribution

    \[ \mathbf{µ}\overline{\mathbf{x}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{312}}{\mathbf{36}}\mathbf{= 8.67}\ \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum fx\overline{}²}}{\mathbf{\sum f}}\mathbf{-}\left( \frac{\mathbf{\sum fx\overline{}}}{\mathbf{\sum f}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{3104}}{\mathbf{36}}\mathbf{-}\left( \frac{\mathbf{312}}{\mathbf{36}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 86.23 - 75.12}\ \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= 11.11}\ \]

    \[ \mathbf{\sigma}\overline{\mathbf{x}}\mathbf{=}\sqrt{\mathbf{11.11}}\mathbf{= \ }\ \]

Population Parameters

X
24
416
864
10100
12144
16256
∑X =52∑X̅²=584

    \[ \mathbf{µ =}\frac{\mathbf{\sum X}}{\mathbf{N}}\mathbf{=}\frac{\mathbf{52}}{\mathbf{6}}\mathbf{= 8.67}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}}{\mathbf{N}}\mathbf{-}\left( \frac{\mathbf{\sum X}}{\mathbf{N}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{584}}{\mathbf{6}}\mathbf{-}\left( \frac{\mathbf{52}}{\mathbf{6}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= 97.33\ -\ 75.11\ }\ \]

    \[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{= \ 22.22}\ \]

    \[ \mathbf{\sigma =}\sqrt{\mathbf{\sigma}^{\mathbf{2}}}\ \]

    \[ \mathbf{\sigma = \ }\sqrt{\mathbf{22.22\ }}\mathbf{\ }\ \]

    \[ \mathbf{\sigma}\mathbf{= \ 4.713}\ \]

Verification

    \[ \mathbf{µ}\overline{\mathbf{x}}\mathbf{= \mu}\ \]

    \[ \mathbf{8.67 = 8.67}\ \]

    \[ \mathbf{\sigma}{\overline{\mathbf{x}}}^{\mathbf{2}}\mathbf{= \ }\frac{\mathbf{\sigma}^{\mathbf{2}}}{\mathbf{n}}\ \]

    \[ \mathbf{11.11 = \ }\frac{\mathbf{22.22}}{\mathbf{2}}\mathbf{= 11.11}\ \]

Q.8 (a) Two samples of sizes 400 and 300 having means 52 and 50 are drawn from the same population of S.D = 3. Test the hypothesis: Ho: µ1 = µ2 Vs H1: µ1>µ2 use α = 0.10.

Solution:

Data

    \[ \mathbf{n}_{\mathbf{1}}\mathbf{= 400,\ }\mathbf{n}_{\mathbf{2}}\mathbf{= 300}\ \]

    \[ \mathbf{\ }{\overline{\mathbf{X}}}_{\mathbf{1}}\mathbf{= 52,\ }{\overline{\mathbf{X}}}_{\mathbf{2}}\mathbf{= 50}\ \]

    \[ \mathbf{\ }\mathbf{S}_{\mathbf{1}}\mathbf{= 3,\ }\mathbf{S}_{\mathbf{2}}\mathbf{= 3,\ \alpha = 0.10}\ \]

    \[ \mathbf{S}_{\mathbf{1}}^{\mathbf{2}}\mathbf{= 9,\ }\mathbf{S}_{\mathbf{2}}^{\mathbf{2}}\mathbf{= 9}\ \]

Step 1: Hypothesis

Null Hypothesis Ho: µ1 = µ2

Alternative Hypothesis : µ1 > µ2

Step 2: Level of Significance α = 0.10

Step 3: Test Statistic

    \[ \mathbf{Z =}\frac{\left( {\overline{\mathbf{X}}}_{\mathbf{1}}\mathbf{-}{\overline{\mathbf{X}}}_{\mathbf{2}} \right)\mathbf{-}\left( \mathbf{\mu}_{\mathbf{1}}\mathbf{-}\mathbf{\mu}_{\mathbf{2}} \right)}{\sqrt{\frac{{\mathbf{s}^{\mathbf{2}}}_{\mathbf{1}}}{\mathbf{n}\mathbf{1}}\mathbf{+}\frac{{\mathbf{s}^{\mathbf{2}}}_{\mathbf{2}}}{\mathbf{n}\mathbf{2}}}}\ \]

Step 4: Critical Region

    \[ \left| \mathbf{Z} \right|\mathbf{> \ 1.289\ \ (}\mathbf{Z}_{\mathbf{\alpha}}\mathbf{=}\mathbf{Z}_{\mathbf{0.10}}\mathbf{= 1.289)}\ \]

Step 5: Computations

    \[ \mathbf{Z =}\frac{\left( {\overline{\mathbf{X}}}_{\mathbf{1}}\mathbf{-}{\overline{\mathbf{X}}}_{\mathbf{2}} \right)\mathbf{-}\left( \mathbf{\mu}_{\mathbf{1}}\mathbf{-}\mathbf{\mu}_{\mathbf{2}} \right)}{\sqrt{\frac{{\mathbf{s}^{\mathbf{2}}}_{\mathbf{1}}}{\mathbf{n}\mathbf{1}}\mathbf{+}\frac{{\mathbf{s}^{\mathbf{2}}}_{\mathbf{2}}}{\mathbf{n}\mathbf{2}}}}\ \]

    \[ \mathbf{Z =}\frac{\left( \mathbf{52 - 50} \right)\mathbf{- 0}}{\sqrt{\frac{\mathbf{9}}{\mathbf{400}}\mathbf{+}\frac{\mathbf{9}}{\mathbf{300}}}}\ \]

    \[ \mathbf{Z =}\frac{\mathbf{2}}{\mathbf{0.2291}}\mathbf{= 8.72}\ \]

Step 6: Conclusion

Our calculated value of Z = 8.72 falls in rejection or critical region so we will reject Null Hypothesis and accept alternative hypothesis. We may conclude that µ1> µ2.

(b) Ten college boys were given a test in statistics. They were given one month’s further tuition and a 2nd test of equal difficulty was held at the end of it. Do the marks given evidence that the students have benefited by the extra coaching? Use α = 0.05

Roll No12345678910
Marks 1st Test48403941384039384339
Marks 2nd Test40394238424340414237

Solution:

Step 1: Hypothesis

Null Hypothesis Ho: μ1=μ2

Alternative HypothesisH1: μ1≠μ2

Step 2: Level of Significance α = 0.05

Step 3: Test Statistic

    \[ \mathbf{t =}\frac{\overline{\mathbf{d}}\mathbf{-}\mathbf{d}_{\mathbf{0}}}{\frac{\mathbf{sd}}{\sqrt{\mathbf{n}}}}\ \]

Step 4: Critical Region

    \[ \mathbf{d.f = n - 1}\ \]

    \[ \mathbf{d.f = 10 - 1 = 9}\ \]

    \[ \mathbf{t}_{\frac{\mathbf{\alpha}}{\mathbf{2}}\mathbf{(d.f)}}\mathbf{=}\mathbf{t}_{\frac{\mathbf{0.05}}{\mathbf{2}}\mathbf{,9}}\mathbf{= 2.262}\ \]

    \[ \mathbf{|t| > \pm}2.262\mathbf{\ }\ \]

Step 5: Computations

    \[ \mathbf{t =}\frac{\overline{\mathbf{d}}\mathbf{-}\mathbf{d}_{\mathbf{0}}}{\frac{\mathbf{sd}}{\sqrt{\mathbf{n}}}}\ \]

X1X2di= X2 – X1di²
4840-864
4039-11
394239
4138-39
3842416
404339
394011
384139
4342-11
3937-24
di=-1 di²=123

    \[ \overline{\mathbf{d}}\mathbf{=}\frac{\mathbf{\sum di}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{- 1}}{\mathbf{9}}\mathbf{= - 0.11}\ \]

    \[ \mathbf{sd =}\sqrt{\left( \frac{\mathbf{1}}{\mathbf{n - 1}}\mathbf{\sum d}\mathbf{i}^{\mathbf{2}}\mathbf{-}\frac{\left( \mathbf{\sum di} \right)^{\mathbf{2}}}{\mathbf{n}} \right)}\ \]

    \[ \mathbf{sd =}\sqrt{\left\lbrack \left( \frac{\mathbf{1}}{\mathbf{9 - 1}} \right)\mathbf{123 -}\frac{\left( \mathbf{- 1} \right)^{\mathbf{2}}}{\mathbf{9}} \right\rbrack}\mathbf{= 3.9068}\ \]

    \[ \mathbf{t =}\frac{\overline{\mathbf{d}}\mathbf{-}\mathbf{d}_{\mathbf{0}}}{\frac{\mathbf{sd}}{\sqrt{\mathbf{n}}}}\ \]

    \[ \mathbf{t =}\frac{\mathbf{- 0.11 - 0}}{\frac{\mathbf{3.9068}}{\sqrt{\mathbf{9}}}}\ \]

    \[ \mathbf{t =}\frac{\mathbf{- 3.04}}{\mathbf{0.8717}}\mathbf{= - \ 0.0844}\ \]

Step 6: Conclusion

Our calculated value of t = -0.0844 falls in acceptance region so we accept Null Hypothesis and reject Alternative Hypothesis. We can conclude that there is no significant evidence to conclude that the extra coaching benefited the students at the α=0.05 level of significance.

Q.9 The following table shows distribution of 400 school children according to physical defect and speech defect α = 0.05. Do the data suggest any association between physical and speech defect?

Speech DefectPhysical Defect
P1P2P3
S1484468
S2422850
S3304842

Solution

Speech DefectPhysical Defect 
P1P2P3Total 
S1484468160 
S2422850120 
S3304842120 
Total120120160400

Solution:

Step 1 Testing the Hypothesis

Ho: There is no Association between Speech Defect & Physical Defect.

H1: There is Association between Speech Defect & Physical Defect.

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(0.05,4)=9.488

The Critical Region χ²cal>9.488

Step 5 Calculation of Expected Frequencies

    \[ \frac{\mathbf{160 \times 120}}{\mathbf{400}}\mathbf{= 48,\ }\frac{\mathbf{160 \times 120}}{\mathbf{400}}\mathbf{= 48,}\frac{\mathbf{160 \times 160}}{\mathbf{400}}\mathbf{= 64\ }\ \]

    \[ \frac{\mathbf{120 \times 120}}{\mathbf{400}}\mathbf{= 36,\ }\frac{\mathbf{120 \times 120}}{\mathbf{400}}\mathbf{= 36,\ }\frac{\mathbf{120 \times 160}}{\mathbf{400}}\mathbf{= 48}\ \]

    \[ \frac{\mathbf{120 \times 120}}{\mathbf{400}}\mathbf{= 36,\ }\frac{\mathbf{120 \times 120}}{\mathbf{400}}\mathbf{= 36,\ }\frac{\mathbf{120 \times 160}}{\mathbf{400}}\mathbf{= 48}\ \]

Step 6 Calculation of χ²:

Computation of χ²
fofefo-fe(fo-fe)²

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

4848000.000
4448-4160.333
68644160.250
42366361.000
2836-8641.778
5048240.083
3036-6361.000
4836121444.000
4248-6360.750
    9.194
    

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


Step 7 Conclusion

Conclusion: The calculated value of χ² is 9.194 is less than the tabulated value of χ² 9.488 or 9.194 falls in the acceptance region. We accept the Null Hypothesis Ho that there is no association between Speech Defect & Physical Defect.

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