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Question 1 (2 marks)
Which one of the following statements is incorrect about the normal
probability distribution?
a. 68.3% of the values of a normal random variable are within ±1
standard deviation of the mean.
b. The larger the value of the standard deviation, the wider and flatter
the curve.
c. The standard normal distribution is symmetric around the
mean of 𝟏𝟏.
d. The 𝑧𝑧-score of the mean of a normal probability distribution is 0.
e. The area to right of mean of a standard normal distribution is 0.5
and the area left of the mean of a standard normal distribution is
also 0.50.
The correct answer is c.
Refer to section 6.2 (properties of a normal distribution and a standard
normal distribution on page 156 of the prescribed textbook.
The standard normal distribution has a mean of 0 and a standard
deviation of 1. We can therefore say it is symmetric around the 0.
Question 2 (2 marks)
Consider a standard normal distribution, 𝑍𝑍∼𝑁𝑁(0.1). Suppose that
the area between −𝑧𝑧 and 𝑧𝑧 is
0.4972 i.e., 𝑃𝑃(−𝑧𝑧≤𝑍𝑍 ≤𝑧𝑧) = 0.4972. Determine the value of 𝑧𝑧 and
choose the correct answer from the list of options below.
a. 0.75
b. 0.25
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c. 𝟎𝟎. 𝟔𝟔𝟔𝟔
d. 0.00
e. 0.50
The correct answer is c.
Given that 𝑃𝑃(−𝑧𝑧≤𝑍𝑍 ≤𝑧𝑧) = 0.4972. The normal distribution is
symmetric, therefore
𝑃𝑃 .
From the above we can write:
𝑃𝑃(𝑍𝑍≤𝑧𝑧) = 0.2486 + 0.5 = 0.7486. From Table 3 of Cumulative
Standardized Normal Probabilities 𝑃𝑃(𝑍𝑍 ≤ 0.67) = 0.7486, the value
is therefore 0.67.
Question 3 (2 marks)
A local company manufactures rechargeable lights with mean battery
life of 80 hours and a standard deviation of 8 hours. Assume that the
battery life is normally distributed. What is the probability that a
randomly selected rechargeable light will have a battery life of at least
68 hours? a. 0.1587
b. 0.8413
c. 0.0668
d. 0.8500
e. 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
The correct answer is e.
Given: 𝑋𝑋 ∼𝑁𝑁(80, 82) i.e., 𝜇𝜇 = 80 hours and 𝜎𝜎 = 8 hours.
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Required: 𝑃𝑃(𝑋𝑋 ≥ 68) =?
𝑋𝑋−𝜇𝜇
We first need to transform 𝑋𝑋 to a standard normal distribution using
the following 𝑍𝑍 = .
𝜎𝜎
𝑋𝑋−𝜇𝜇68 − 80
𝑃𝑃(𝑋𝑋 ≥ 68) = 𝑃𝑃 𝜎 8 ≥
= 𝑃𝑃(𝑍𝑍≥−1.5) 𝜎
= 𝑃𝑃(𝑍𝑍≤ 1.5)
= 0.9332
Alternatively: 𝑃𝑃(𝑋𝑋 ≥−1.5) = 1 −𝑃𝑃(𝑍𝑍
< −1.5)
= 1 − 0.0668
= 0.9332
Question 4 (2 marks)
Consider the information from the previous question. Remember
rechargeable lights have a mean battery life of 80 hours with a standard
deviation of 8 hours. Suppose that the local company manufactures the
lights for a lighting warehouse. It is observed that the warehouse rejects
about 20% of the lights (because they do not pass the minimum battery
life test). Determine the minimum battery life accepted by the lighting
warehouse
a. Approximately 72 hours.
b. Approximately 75 hours.
c. Approximately 𝟔𝟔𝟗𝟗 hours.
d. Approximately 76 hours.
e. Approximately 74 hours.
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Question 1 (2 marks)
Which one of the following statements is incorrect about the normal
probability distribution?
a. 68.3% of the values of a normal random variable are within ±1
standard deviation of the mean.
b. The larger the value of the standard deviation, the wider and flatter
the curve.
c. The standard normal distribution is symmetric around the
mean of 𝟏𝟏.
d. The 𝑧𝑧-score of the mean of a normal probability distribution is 0.
e. The area to right of mean of a standard normal distribution is 0.5
and the area left of the mean of a standard normal distribution is
also 0.50.
The correct answer is c.
Refer to section 6.2 (properties of a normal distribution and a standard
normal distribution on page 156 of the prescribed textbook.
The standard normal distribution has a mean of 0 and a standard
deviation of 1. We can therefore say it is symmetric around the 0.
Question 2 (2 marks)
Consider a standard normal distribution, 𝑍𝑍∼𝑁𝑁(0.1). Suppose that
the area between −𝑧𝑧 and 𝑧𝑧 is
0.4972 i.e., 𝑃𝑃(−𝑧𝑧≤𝑍𝑍 ≤𝑧𝑧) = 0.4972. Determine the value of 𝑧𝑧 and
choose the correct answer from the list of options below.
a. 0.75
b. 0.25
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c. 𝟎𝟎. 𝟔𝟔𝟔𝟔
d. 0.00
e. 0.50
The correct answer is c.
Given that 𝑃𝑃(−𝑧𝑧≤𝑍𝑍 ≤𝑧𝑧) = 0.4972. The normal distribution is
symmetric, therefore
𝑃𝑃 .
From the above we can write:
𝑃𝑃(𝑍𝑍≤𝑧𝑧) = 0.2486 + 0.5 = 0.7486. From Table 3 of Cumulative
Standardized Normal Probabilities 𝑃𝑃(𝑍𝑍 ≤ 0.67) = 0.7486, the value
is therefore 0.67.
Question 3 (2 marks)
A local company manufactures rechargeable lights with mean battery
life of 80 hours and a standard deviation of 8 hours. Assume that the
battery life is normally distributed. What is the probability that a
randomly selected rechargeable light will have a battery life of at least
68 hours? a. 0.1587
b. 0.8413
c. 0.0668
d. 0.8500
e. 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
The correct answer is e.
Given: 𝑋𝑋 ∼𝑁𝑁(80, 82) i.e., 𝜇𝜇 = 80 hours and 𝜎𝜎 = 8 hours.
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Required: 𝑃𝑃(𝑋𝑋 ≥ 68) =?
𝑋𝑋−𝜇𝜇
We first need to transform 𝑋𝑋 to a standard normal distribution using
the following 𝑍𝑍 = .
𝜎𝜎
𝑋𝑋−𝜇𝜇68 − 80
𝑃𝑃(𝑋𝑋 ≥ 68) = 𝑃𝑃 𝜎 8 ≥
= 𝑃𝑃(𝑍𝑍≥−1.5) 𝜎
= 𝑃𝑃(𝑍𝑍≤ 1.5)
= 0.9332
Alternatively: 𝑃𝑃(𝑋𝑋 ≥−1.5) = 1 −𝑃𝑃(𝑍𝑍
< −1.5)
= 1 − 0.0668
= 0.9332
Question 4 (2 marks)
Consider the information from the previous question. Remember
rechargeable lights have a mean battery life of 80 hours with a standard
deviation of 8 hours. Suppose that the local company manufactures the
lights for a lighting warehouse. It is observed that the warehouse rejects
about 20% of the lights (because they do not pass the minimum battery
life test). Determine the minimum battery life accepted by the lighting
warehouse
a. Approximately 72 hours.
b. Approximately 75 hours.
c. Approximately 𝟔𝟔𝟗𝟗 hours.
d. Approximately 76 hours.
e. Approximately 74 hours.
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