Answer the following questions:

Questions 1 - 5: Consider the following cumulative probability distribution table of a random variable
X representing the number of visitors in an hour:

𝑥 0 1 2 3 4 5
𝑃(𝑋 ≤ 𝑥) 0.2 0.3 0.4 0.6 0.9 k
1
1. The value of 𝑘 equals to
A) 1 B) 0.2 C) 0.1 D) 0.4

This is the cumulative distribution function, therefore the value of 𝑘 at the last probability value
equals 1. The probability density function 𝑓(𝑥):
𝑥 0 1 2 3 4 5
𝐹(𝑥) 0.2 0.3 0.4 0.6 0.9 1
𝑓(𝑥) 0.2 0.1 0.1 0.2 0.3 0.1

2. The probability that at most one visitor will attend the store in a given hour is
1
A) 0.1 B) 0.4 C) 0.6 D) 0.3

Using 𝐹(𝑥):

𝑃(𝑋 ≤ 1) = 𝐹(1) = 0.3

Using 𝑓(𝑥):

𝑃(𝑋 ≤ 1) = 𝑃(1) + 𝑓(0) = 0.5

3. P( 3 ≤ X < 5) =
A) 0.6 B) 0.5 C) 0.7 D) 0.9

𝑃(3 ≤ 𝑋 < 5) = 𝑃(𝑋 < 5) − 𝑃(𝑋 < 3) = 𝑃(𝑋 ≤ 4) − 𝑃(𝑋 ≤ 2) = 𝐹(4) − 𝐹(2)

= 0.9 − 0.4 = 0.5

4. The mean number of visitors who attend a store in one hour is
A) 2.0 B) 1.5 C) 2.6 D) 0.43
𝑥 0 1 2 3 4 5
𝐹(𝑥) 0.2 0.3 0.4 0.6 0.9 1
𝑓(𝑥) 0.2 0.1 0.1 0.2 0.3 0.1

𝐸(𝑋) = 1 0.3) + (5 × 0.1)

𝑥𝑓(𝑥) = (0 × 0.2) + (1 × 0.1) + (2 × 0.1) + (3 × 0.2) + (4 ×

𝐸(𝑋) = 2.6

5. The variance of the number of visitors who attend a store in one hour is
A) 2.84 B) 0.1 C) 2.6 D) 1.685
𝑥 0 1 2 3 4 5
𝐹(𝑥) 0.2 0.3 0.4 0.6 0.9 1
𝑓(𝑥) 0.2 0.1 0.1 0.2 0.3 0.1

𝐸(𝑋) = 1 0.3) + (5 × 0.1)

𝑥𝑓(𝑥) = (0 × 0.2) + (1 × 0.1) + (2 × 0.1) + (3 × 0.2) + (4 ×

𝐸(𝑋) = 2.6

𝐸(𝑋 ) = 𝑥 𝑓(𝑥)

= (0 × 0.2) + (1 × 0.1) + (2 × 0.1) + (3 × 0.2) + (4 × 0.3) + (5 × 0.1)

𝐸(𝑋 ) = 9.6

𝑉𝑎𝑟(𝑋) = 𝐸(𝑋 ) − [𝐸(𝑋)] = 9.6 − 2.6 = 2.84

Questions 6 - 8: Let Z be a standard normal N(0,1), then:

6. The area under the curve which lies to the left of z = 1.39:
A) 0.0823 B) 0.0048 C) 0.71904 D) 0.0015

𝑃(𝑍 < −1.39) = 0.08226 ≈ 0.0823

7. The probability P(Z > 1.32) equals to:
A) 1 B) 0 C) 0.00371 D) 0.0934

𝑃(𝑍 > 1.32) = 1 − 𝑃(𝑋 < 1.32) = 1 − 0.9066 = 0.0934

8. The value of k such that P(0.93 < Z < k) = 0.0427 equals to:
A) 0.8665 B) -1.11 C) 1.11 D) 1

𝑃(0.93 < 𝑍 < 𝑘) = 0.0427

𝑃(𝑍 < 𝑘) − 𝑃(𝑍 < 0.93) = 0.0427 → 𝑃(𝑍 < 𝑘) = 0.0427 + 0.8238

→ 𝑃(𝑍 < 𝑘) = 0.8665 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑡𝑎𝑏𝑙𝑒 → 𝑘 = 1.11

Questions 9 - 13: Suppose that the probability that a Saudi man has high blood pressure is 0.15.
Suppose that we randomly select a sample of 12 Saudi men. Suppose that X the number of men
with high blood pressure in the sample

9. The distribution of the random variable X is:

A) B) C)
0
D)
Poisson(12) Normal(0,1) Normal(12,0.15) Binomial(12,0.15)

This is the Binomial distribution with 𝑝 = 0.15 𝑛 = 12

10. P(X ≤ 1) is equal to.

A) 0.5 B) 0.443 C) 0.98 D) 0.014

𝑃(𝑋 ≤ 1) = 𝑓(0) + 𝑓(1) = 0.4434

11. P(X > 1) is equal to.

A) 0.557 B) 0.004 C) 0.789 D) 0.548

𝑃(𝑋 > 1) = 1 − 𝑃(𝑋 ≤ 1) = 1 − 0.443 = 0.557

12. The mean of X is equal to:

A) 1.8 B) 0.2 C) 10 D) 3

𝐸(𝑥) = 𝑛𝑝 = 12 × 0.15 = 1.8

13. The standard deviation of X is equal to:
A) 0 B) 1 C) -1 D) 1.24

𝜎 = 𝑉𝑎𝑟(𝑋) = 𝑛𝑝𝑞 = 12 × 0.15 × 0.85 = 1.53

𝜎 = √1.53 = 1.24

Questions 14 - 15: The weight X of newborn children in a given population is normally distributed
with a mean 3.25 kilograms and a standard deviation of 0.9 kilograms. If we select a sample of 30
children, then:

14. The mean of the sample mean is equal to:

A) 2.5 B) 3.25 C) 30 D) 97.5

𝑋~𝑁(3.25,0.9) 𝑛 = 30

𝑋 = 𝜇 = 3.25

15. The standard error of the sample mean is equal to:

A) 0.456 B) 0.201 C) 0.164 D) 0.324

𝜎 0.9
𝑆𝐸 = = = 1.64
√𝑛 √30
Questions 16 - 17
16. Let X has normal distribution with mean = µ and standard deviation = σ, then
P(X < µ) =
A) 0.5 B) 0.0 C) 0.1 D) 1.0
𝜇−𝜇
𝑃(𝑋 < 𝜇) = 𝑃 𝑍 < = 𝑃(𝑍 < 0) = 0.5
𝜎

17. Let X has normal distribution with mean µ = 0 and standard deviation σ, the value of σ such
that
P(X < 0.5) = 0.95 is equal to
A) 0.50 B) 0.304 C) 0.1.645 D) 0.95

0.5 − 𝜇
𝑃(𝑋 < 0.5) = 0.95 → 𝑃 𝑍 < = 0.95
𝜎
0.5 − 𝜇 0.5 0.5
→ = = 1.65 → 𝜎 = = 0.3030
𝜎 𝜎 1.65
Questions 18 – 19 Let T follow the t-distribution with 14 degrees of freedom, then:

18. The t-value that leaves an area of 0.95 to the right is equal to:
A) 2.5 B) -1.761 C) 4.2 D) 1.341

19. The t-value that leaves an area of 0.95 to the left is equal to:
A) 1.761 B) 2.795 C) 1.247 D) 9.515

Questions 20 – 22 Suppose that in a certain city, the weekly number of infected cases with Corona
virus (COVID-19) has a Poisson distribution with an average (mean) of 5 cases per week.

20. The probability that there will be no infected cases this week equals to:
A) 0.9933 B) 0.8779 C) 0.0067 D) 0.4567

𝑋~𝑃𝑜𝑖𝑠𝑠𝑜𝑛(𝜆 = 5) 𝑒𝑣𝑒𝑟𝑦 𝑤𝑒𝑒𝑘

𝑒 5
𝑃(𝑋 = 0) = 𝑓(0) = = 0.0067
0!

21. The probability that there will be 2 infected cases within two weeks equals to:
A) 0.301 B) 0.8750 C) 0.0023 D) 0.998

𝑋~𝑃𝑜𝑖𝑠𝑠𝑜𝑛(𝜆 = 10) 𝑒𝑣𝑒𝑟𝑦 𝑡𝑤𝑜 𝑤𝑒𝑒𝑘

𝑒 10
𝑃(𝑋 = 2) = 𝑓(2) = = 0.0023
2!

22. The variance of the number of infected cases with Corona virus (COVID-19) infected cases in
two weeks equal to
A) 10 B) 25 C) 5 D) 1

𝑋~𝑃𝑜𝑖𝑠𝑠𝑜𝑛(𝜆 = 10) 𝑒𝑣𝑒𝑟𝑦 𝑡𝑤𝑜 𝑤𝑒𝑒𝑘

𝑉𝑎𝑟(𝑋) = 𝜆 = 10

I
Questions 23 - 24: Suppose that 10% of the students in a certain university smoke cigarette. A
random sample of 90 student is taken from this university. Let 𝑝̂ be the proportion of smokers in
the sample.
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23. The distribution of 𝑝̂ is: Valool
A) Binomial(30,0.1) B) Normal(0.1,0.001) C) Normal(0,1) D) Binomial(30,0.9)

𝑝𝑞 0.1 × 0.9
𝑝̂ ~𝑁 𝑝 = 0.1, = = √0.001
𝑛 90

24. 𝑃(𝑝̂ > 0.15) equals to:

A) 0.41245 B) 0.99614 C) 0.057 D) 0.005

0.15 − 0.1

1
PREET
𝑃(𝑝̂ > 0.15) = 1 − 𝑃 𝑍 <
√0.001

= 1 − 𝑃(𝑍 < 1.58) = 1 − 0.9429 = 0.057

P 2 1198 1 9429
1,1
Questions 25 - 27:

In
Suppose that 40% of residents in population A have medical insurance and 30% of residents in
population B have medical insurance. We have randomly and independently selected a sample of
130 from population A and another sample of 120 from population B. Answer the following.
hi_ inn
25. The value of 𝜇 =⋯
A) 0.01 B) 0.1 C) 0.03 D) 0.3

𝑝 𝑞 𝑝 𝑞 0.4 × 0.6 0.3 × 0.7

𝑝̂ − 𝑝̂ ~𝑁 𝑝 − 𝑝 = 0.4 − 0.3 = 0.1, + = +
𝑛 𝑛 130 120

𝜇 = 0.1

26. The value of 𝜎

A) 0.0056
=⋯
B) 0.0046 C) 0.0036
Pint
D)
Pit
0.0026

𝑝 𝑞 𝑝 𝑞 0.4 × 0.6 0.3 × 0.7

𝜎 = + = + = 0.0035
𝑛 𝑛 130 120

27. 𝑃(0.05 ≤ 𝑝̂ − 𝑝̂ ≤ 0.2) = ⋯

A) 0.7081 B) 0.7882 C) 0.7728 D) 0.750

0.2 − 0.1 0.05 − 0.1

𝑃(0.05 ≤ 𝑝̂ − 𝑝̂ ≤ 0.2) = 𝑃 𝑍 ≤ −𝑃 𝑍 ≤
√0.0035 √0.0035

Pla f PEE
 1,66 22 012
P 2 0

p
=𝑃 𝑍≤
9515
0.2 − 0.1
−𝑃 𝑍 ≤
0.05 − 0.1 012033
= 𝑃(𝑍 ≤ 1.67) − 𝑃(𝑍 ≤ −0.83)
017
√0.0035 √0.0035

= 0.9525 − 0.2033 = 0.7492 ≈ 0.750

Questions 28 - 30:
In a study carried out on U.S. females, it was found that serum cholesterol levels are given as
follows:
x̅
Population Mean Standard deviation
A 185 37.2
B 195 34.7

I
Suppose you select a simple random sample of size 50 independently from each population. Then:

28. The distribution of (𝑋 − 𝑋 ) is:

A) Normal(10,50) B) Normal(0,1) C) Normal(10, 36.45)
0
D) Normal(10, 51.76)

𝜎 𝜎 34.7 37.2
𝑋 − 𝑋 ~𝑁 𝜇 − 𝜇 = 10, + = + = √51.76
𝑛 𝑛 50 50

29. The probability that the difference between sample means (𝑋 − 𝑋 ) will be more than 8?
A) 0.8397 B) 0.609 C) 0.3779 D) 0.8297

8 − 10
𝑃(𝑋 − 𝑋 > 8) = 1 − 𝑃 𝑍 ≤ = 1 − 𝑃(𝑍 ≤ −0.27)
50.76
Xp A 81 1 P 22
1
1 − 0.3936 = 0.609i P 22 927
30. 𝑃((𝑋 − 𝑋 ) > 10) equals to:
1 012931
A) 0.90012 B) 0.0415 C) 0.50 D) 0.67771

10 − 10
𝑃(𝑋 − 𝑋 > 10) = 1 − 𝑃 𝑍 ≤
50.76
= 1 − 𝑃(𝑍 ≤ 0) = 1 − 0.5 = 0.5

1120 as
1 P as 1
0 0
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