Statistics: Sample Test Solution
Chapters 5: Discrete Probability Distribution
1. Does the table describe probability distribution? What is the random variable, what are its
possiblevalues, and are its values numerical?
Number of Girls in 3 Births
Number of girls x
P(x)
0
0.125
1
0.375
2
0.375
3
0.125
Answer:
Yes, there are 3 criteria:
1) The values of the random value x are numerical.
2) Their sum: 0.125 + 0.125 + 0.375+0.375 =1
3) The values of the random value x are between 0 and 1.
The random variable is x, which is the number of girls in three births.
The possible values of x are0, 1,2, and 3.
The values of the random value x are numerical.
2. In a game, you pay 60 cents to select a 4-digit number. If you win by selecting the correct
4-digit number, you collect $3,000.
a) How many different selections are possible?
n = number of digits = 10:
0, 1, 2,…,9
10 possible numbers in 4 places (numbers can repeat)
104 = 10,000
b) What is the probability of winning?
P(w) = n(w) / n(S) = 1/10,000
c) If you win, what is your net profit?
net profit = money gained – investment (Cost) = 3000 – 0.60 = $2999.40
d) Write the Probability Distribution of Net Profit if you win.
Event
Lose
X
-0.60
Win
+2999.40
P(x)
9,999/10,000
(0.9999)
1/10,000 (0.0001)
1
x P(x)
-0.59994
0.29994
Note: Last column is not part of the Probability Distribution; it is used to calculate the
expected values in the next question.
e) Find the expected value and interpret.
E = μ = ∑(x P(x)) = -0.6 x 0.9999 + 2999.4 x 0.0001 = -0.3
Anyone who plays the game loses about 30 cents on the average.
3. A pharmaceutical company receives large shipments of aspirin tablets. The acceptance
sampling plan is to randomly select and test 24 tablets. The entire shipment is accepted if
at most 2 tablets do not meet the required specifications. If a particular shipment of
thousands of aspirin tablets actually has a 5.0% rate of defects, what is the probability
that this whole shipment will be accepted?
Answer:
( )
(
)
Given:
P(x ≤ 2) (x = 0, 1, 2) = P (0) + P (1) + P (2) = 0.884
4. It is known that 90% of managers of all companies suffer from job related stress. What is
the probability that in a sample of 10 managers,
a) Exactly 8 suffer from job related stress.
( )
(
)
Given:
( )
b) At most 2 suffer from job related stress.
P(x ≤ 2) (x = 0,1,2)
= P(0) + P(1) + P(2)
P(0) = 10!/0! 10! X (0.9)0 x (0.1)10 = 1 x 1 x 0.110
P(1) = 10!/1! 9! X 0.91 x (0.1)9 = 10 x 0.9 x 0.19 = 0.000000009
P(2)= 10!/2! 8! X (0.9)2 x (0.1)8 = 45 x (0.9)2 x (0.1)8 = 0.0000003645
P(x ≤ 2) = P(0) + P(1) + P(2) = 0.0000003746 or 3.746 x 10-7
c) At least 3 suffer from job related stress.
x ≥3 (x=3,4,5,6,7,8,9,10)
P(x≥3) = P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) = 1- P(x ≤2)
= 1 –[ (P(0) + P(1) + P(2)] = =1 – 0.0000003746 = 0.999999625
2
d) Find the expected value.
E=μ=nxp
E = 10 x 0.9 = 9
e) Find the standard deviation.
σ2 = npq
σ2 = 10( 0.9)( 0.1) = 0.9
σ=√
σ = 0.9487
f) Would it be unusual to claim that at least 7 managers from this sample suffer from job
related stress?
Range rule of thumb: μ ± 2σ
9 ± 2(0.9487)
usual minimum: 7.1026
usual maximum: 10. 8974
7  (7.1026,10.8974)  Unusual
5. Find the probability of a couple having at least one girl among 3 children.
Method 1:
P (at least 1) = 1 – P (None) = 1 – (1/2)3 = 7/8, Notice: p(g) = p(b) = 1/2
Method 2:
Binomial Distribution: n = 3, x ≥1, p = q = ½
x = 1, 2, 3
P(1) + P(2) + P(3) = 1 – P(0)
( )
(
)
P(0) = 3! /0! 3! X (0.5)0 x (0.5)3
P (0) = 1 x (0.5)3
P(x ≥ 1) = 1 – 0.53 = 1 – 0.125 = 0.875 (or 7/8)
6. If an alarm clock has a 0.9 probability of working on any given morning.
a) What is the probability that it will not work?
( ̅)
( )
1 – 0.9 = 0.1 (or 10 percent)
b) What is the probability that 2 such alarm clocks will not work?
(̅̅̅̅ ̅̅̅̅)
(̅̅̅̅) (̅̅̅̅)
0.1 x 0.1 = 0.01
there is the probability of 1% that 2 such alarm clocks will not work.
c) What is the probability of being awakened if you have 2 such alarm clocks?
Method 1:
3
Probability of being awakened = 1 – probability of not being awakened
1 – 0.01 = 0.99
the probability of being awakened if you have 2 such alarm clocks is 99%.
Method 2:
P(A U B) = P(A) + P(B) – P(A∩B) = 0.9 + 0.9 – (0.9 x 0.9) = 0.99
Method 3:
Binomial Distribution: n = 2, p = 0.9
( )
Find the probability of: P(x ≥1) = 1 – P (0)
( )
(
)
4
(
)