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 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.
Table of Contents
Statistics II Solved Paper 2007 FBISE
MCQS
Fill the relevant bubble against each question according to curriculum: | |||
1 | A and B are independent events if P(A/B) = P(B) | ||
A) | True | B) | False |
2 | P(A) = ¾ means that odds are 3 to 4 that event A will occur: | ||
A) | True | B) | False |
3 | Probability of drawing a red card from a pack of 52 cards is 26/52: | ||
A) | True | B) | False |
4 | For two independent variables X and Y Var(X – Y) = Var(X) – Var(Y): | ||
A) | True | B) | False |
5 | E(6X + 15) = 6E(X)+….. | ||
A) | True | B) | False |
6 | A binomial distribution will be symmetrical if p and q are: | ||
A) | Different | B) | Distinct |
C) | Identical | D) | None of these |
7 | For a binomial distribution ∑(q+p) = 1 | ||
A) | True | B) | False |
8 | Area under standard normal curve is….. | ||
A) | 1 | B) | 0 |
C) | 100 | D) | 10 |
9 | Normal distribution and Gaussian distribution are two different distributions: | ||
A) | True | B) | False |
10 | The factor is called the finite population correction factor. | ||
A) | n/N | B) | |
C) | N/n | D) | None of these |
11 | Sampling error is the difference between the population parameter and the | ||
A) | parameter | B) | sample statistic |
C) | sample | D) | None of these |
12 | For α = 0.01 the critical value of z = …. For two tailed test. | ||
A) | 2.58 | B) | 1.96 |
C) | 2.33 | D) | 5.96 |
13 | If a deserving player is not selected in the team it is…. | ||
A) | Type I Error | B) | Type II Error |
C) | Correct Decision | D) | None of these |
14 | If (AB) = [(A)(B)]/N then A and B are said to be independent. | ||
A) | True | B) | False |
15 | Ho: µ ≥ 30 is a …… hypothesis. | ||
A) | One sided | B) | Two sided |
C) | Alternative | D) | None of these |

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)
(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.
- Total Probability equals to one
Total area of PDF is equals to one in which we assume that random variable will take some value.
- Probability of single value is zero
In PDF, the probability of a single value is zero.
- 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.
(iv) If a variable has the binomial distribution determine its Variance.
Solution
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:
- The mean, median and mode are equal.
- Mean deviation is 0.7979 of its standard deviation.
- Quartile Deviation is 0.6745 of its standard deviation.
- The curve is symmetrical about ordinate at X̅ = µ.
- 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, 1 | 2, 1 | 3, 1 | 4, 1 | 5, 1 | 6, 1 |
1, 2 | 2, 2 | 3, 2 | 4, 2 | 5, 2 | 6, 2 |
1, 3 | 2, 3 | 3, 3 | 4, 3 | 5, 3 | 6, 3 |
1, 4 | 2, 4 | 3, 4 | 4, 4 | 5, 4 | 6, 4 |
1, 5 | 2, 5 | 3, 5 | 4, 5 | 5, 5 | 6, 5 |
1, 6 | 2, 6 | 3, 6 | 4, 6 | 5, 6 | 6, 6 |
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
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
Events
X | P(X) | xp(x) | x²p(x) |
1 | 7/462 | 7/462 | 7/462 |
2 | 84/462 | 168/462 | 336/462 |
3 | 210/462 | 630/462 | 1890/462 |
4 | 140/462 | 560/462 | 2240/462 |
5 | 21/462 | 105/462 | 525/462 |
∑xp(x) = 1470/462 | ∑x²p(x) = 4998/462 |
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:
Here N = 1024, n = 5, x = 1 and 2
Dividing equation ii by equation i
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)
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)
Solution (iii)
Here µ = 47.6, = 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
Solution (iv)
Here µ = 47.6, = 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
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:
SolutionPopulation = 2, 4, 8, 10, 12, 16, Population Size N = 6, Sample size n = 2
S/No | Samples | Sum of Samples | Mean of Samples | S/No | Samples | Sum of Samples | Mean of Samples |
1 | 2, 2 | 4 | 2 | 19 | 10, 2 | 12 | 6 |
2 | 2, 4 | 6 | 3 | 20 | 10, 4 | 14 | 7 |
3 | 2, 8 | 10 | 5 | 21 | 10, 8 | 18 | 9 |
4 | 2, 10 | 12 | 6 | 22 | 10, 10 | 20 | 10 |
5 | 2, 12 | 14 | 7 | 23 | 10, 12 | 22 | 11 |
6 | 2, 16 | 18 | 9 | 24 | 10, 16 | 26 | 13 |
7 | 4, 2 | 6 | 3 | 25 | 12, 2 | 14 | 7 |
8 | 4, 4 | 8 | 4 | 26 | 12, 4 | 16 | 8 |
9 | 4, 8 | 12 | 6 | 27 | 12, 8 | 20 | 10 |
10 | 4, 10 | 14 | 7 | 28 | 12, 10 | 22 | 11 |
11 | 4, 12 | 16 | 8 | 29 | 12, 12 | 24 | 12 |
12 | 4, 16 | 20 | 10 | 30 | 12, 16 | 28 | 14 |
13 | 8, 2 | 10 | 5 | 31 | 16, 2 | 18 | 9 |
14 | 8, 4 | 12 | 6 | 32 | 16, 4 | 20 | 10 |
15 | 8, 8 | 16 | 8 | 33 | 16, 8 | 24 | 12 |
16 | 8, 10 | 18 | 9 | 34 | 16, 10 | 26 | 13 |
17 | 8, 12 | 20 | 10 | 35 | 16, 12 | 28 | 14 |
18 | 8, 16 | 24 | 12 | 36 | 16, 16 | 32 | 16 |
Sampling Distribution of X̅ | |||
X̅ | f | fX̅ | fX̅² |
2 | 1 | 2 | 4 |
3 | 2 | 6 | 18 |
4 | 1 | 4 | 16 |
5 | 2 | 10 | 50 |
6 | 4 | 24 | 144 |
7 | 4 | 28 | 196 |
8 | 3 | 24 | 192 |
9 | 4 | 36 | 324 |
10 | 5 | 50 | 500 |
11 | 2 | 22 | 242 |
12 | 3 | 36 | 432 |
13 | 2 | 26 | 338 |
14 | 2 | 28 | 392 |
16 | 1 | 16 | 256 |
Sum | 36 | 312 | 3104 |
∑f = | ∑fX̅ = | ∑fX̅²= |
Mean, Variance & S.D of Sampling Distribution
Population Parameters
X | X² |
2 | 4 |
4 | 16 |
8 | 64 |
10 | 100 |
12 | 144 |
16 | 256 |
∑X =52 | ∑X̅²=584 |
Verification
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
Step 1: Hypothesis
Null Hypothesis Ho: µ1 = µ2
Alternative Hypothesis : µ1 > µ2
Step 2: Level of Significance α = 0.10
Step 3: Test Statistic
Step 4: Critical Region
Step 5: Computations
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 No | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Marks 1st Test | 48 | 40 | 39 | 41 | 38 | 40 | 39 | 38 | 43 | 39 |
Marks 2nd Test | 40 | 39 | 42 | 38 | 42 | 43 | 40 | 41 | 42 | 37 |
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
Step 4: Critical Region
Step 5: Computations
X1 | X2 | di= X2 – X1 | di² |
48 | 40 | -8 | 64 |
40 | 39 | -1 | 1 |
39 | 42 | 3 | 9 |
41 | 38 | -3 | 9 |
38 | 42 | 4 | 16 |
40 | 43 | 3 | 9 |
39 | 40 | 1 | 1 |
38 | 41 | 3 | 9 |
43 | 42 | -1 | 1 |
39 | 37 | -2 | 4 |
∑ di=-1 | ∑ di²=123 |
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 Defect | Physical Defect | ||
P1 | P2 | P3 | |
S1 | 48 | 44 | 68 |
S2 | 42 | 28 | 50 |
S3 | 30 | 48 | 42 |
Solution
Speech Defect | Physical Defect | ||||
P1 | P2 | P3 | Total | ||
S1 | 48 | 44 | 68 | 160 | |
S2 | 42 | 28 | 50 | 120 | |
S3 | 30 | 48 | 42 | 120 | |
Total | 120 | 120 | 160 | 400 |
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
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
Step 6 Calculation of χ²:
Computation of χ² | ||||
fo | fe | fo-fe | (fo-fe)² | |
48 | 48 | 0 | 0 | 0.000 |
44 | 48 | -4 | 16 | 0.333 |
68 | 64 | 4 | 16 | 0.250 |
42 | 36 | 6 | 36 | 1.000 |
28 | 36 | -8 | 64 | 1.778 |
50 | 48 | 2 | 4 | 0.083 |
30 | 36 | -6 | 36 | 1.000 |
48 | 36 | 12 | 144 | 4.000 |
42 | 48 | -6 | 36 | 0.750 |
9.194 | ||||
|
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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