**This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “F-Distribution”.**

**1. The mean of the f – distribution is equal to ___________**a) v

_{2}/ (v

_{2}– 2) for v

_{2}> 2

b) v

_{2}/ (v

_{2}– 2)

^{2}for v

_{2}> 2

c) v

_{2}/ (v

_{2}– 2)

^{3}for v

_{2}> 2

d) v

_{2}/ (v

_{2}– 2)

^{-1}for v

_{2}> 2

**Answer: a**Explanation: The mean of the distribution is equal to v

_{2}/ (v

_{2}– 2) for v

_{2}> 2. v

_{2}denotes the degree of freedom of F-Distribution.

**2. Variance is equal to [(v _{1} + v_{2} – 2)] / [v_{1} * (v_{2} – 2)^{2} * (v_{2} – 4)] for v_{2} > 4 for a f-Distribution.**a) True

b) False

**Answer: b**Explanation: Variance is equal to [2 * \(v_2^2\) * (v

_{1}+ v

_{2}– 2)] / [ v

_{1}* (v

_{2}– 2)

^{2}* (v

_{2}– 4)] for v

_{2}> 4 for a f-Distribution where v

_{1}and v

_{2}denote the degrees of freedom of f-Distribution.

**3. Which of the following distributions is Continuous?**a) Binomial Distribution

b) Hyper-geometric Distribution

c) F-Distribution

d) Poisson Distribution

**Answer: c**Explanation: Binomial, Poisson and Hyper geometric distributions are Discrete Distributions. Only F- Distribution is Continuous Distribution in the given Distributions.

**4. Which of the following distributions is used to compare two variances?**a) T – Distribution

b) F – Distribution

c) Normal Distribution

d) Poisson Distribution

**Answer: b**Explanation: F – Distribution is used when we require for comparing two variances. It uses a f-Test to compare two values of variances.

**5. F-Distribution cannot take negative values.**a) True

b) False

**Answer: a**Explanation: The value of the F-distribution is always positive, or zero. The variances are the square of the deviations and hence cannot assume negative values. Its value lies between 0 and ∞.

**6. Find Variance for an F-Distribution with v _{1}=5 and v_{2}=9.**a) 1.587

b) 1.378

c) 1.578

d) 1.498

**Answer: a**Explanation: For a f – Distribution:

Var(X) = [2 * \(v_2^2\) * (v

_{1}+ v

_{2}– 2)] / [v

_{1}* (v

_{2}– 2)

^{2}* (v

_{2}– 4)] for v

_{2}> 4 where v

_{1}and v

_{2}denote the degrees of freedom of f-Distribution.

Hence Var(X) = [2 * 9

^{2}* (14 – 2)] / [5 * (9 – 2)

^{2}* (9- 4)] = 1.587.

**7. The table shows the standard Deviation and Sample Standard Deviation for both men and women. Find the f statistic considering the Men population in numerator.**

Population | Population Standard Deviation | Sample Standard Deviation |

Men | 30 | 35 |

Women | 50 | 45 |

a) 2.68

b) 1.34

c) 1.68

d) 1.34

**Answer: c**Explanation: The f -statistic is calculated using the following equation:

f = [s

_{1}

^{2}/σ

_{1}

^{2}]/[s

_{2}

^{2}/σ

_{2}

^{2}].

where σ

_{1}is the standard deviation of population 1

s

_{1}is the standard deviation of the sample drawn from population 1

σ

_{2}is the standard deviation of population 2

s

_{1}is the standard deviation of the sample drawn from population 2.

f = (35

^{2}/30

^{2})/(45

^{2}/50

^{2})

f = (1225/900)/(2025/2500)

f = 1.361/0.81 = 1.68.

**8. Calculate the value of f-statistic having a cumulative probability of 0.95.**a) 0.55

b) 0.5

c) 0.05

d) 0.05

**Answer: d**Explanation: The relation between f – statistic and cumulative probability is given as

If f – statistic = f α then, cumulative probability = (1 – α)

Hence, for cumulative probability 0.95

f – statistic = (1 – 0.95) = 0.05.

**9. There is only 1 parameter in F-Distribution.**a) True

b) False

**Answer: b**Explanation: There are 2 parameters in F-Distribution v

_{1}and v

_{2}. They are called degrees of freedom of F-Distribution.

**10. Find the Expectation for a F- Distribution variable with v _{1} = 7 and v_{2} = 8.**a) 4/7

b) 4/6

c) 4/3

d) 4/5

**Answer: c**Explanation: The Expectation for F-Distribution is given as

E(X) = v

_{2}/ (v

_{2}– 2) for v

_{2}> 2

Hence, E(X) = 8 / (8-2)

E(X) = 4/3.