STATS 250
Assignment # 7
Full credit
Extra credit
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Antonín Koukolík
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241/19
The average amount of precipitation in Dallas, Texas, during the month of April is 3.5 inches (The World
Almanac, 2000). Assume that a normal distribution applies and that the standard deviation is .8 inches.
a. What percentage of the time does the amount of rainfall in April exceed 5 inches? (1 point)
5 − 3.5
) = 𝑃(𝑧 > 1.875) = 0.0304
0.8
Therefore it exceeds 5 inches approximately 3.04% of the time.
𝑃(𝑥 > 5) = 𝑃 (𝑧 >
b. What percentage of the time is the amount of rainfall in April less than 3 inches? (1 point)
3 − 3.5
) = 𝑃(𝑧 < −0.625) = 0.2660
0.8
Therefore it is less than 3 inches 26.6% of the time.
𝑃(𝑥 < 3) = 𝑃 (𝑧 <
c. A month is classified as extremely wet if the amount of rainfall is in the upper 10% for that
month. How much precipitation must fall in April for it to be classified as extremely wet? (1
point)
𝑥 = 𝑧𝜎 + 𝜇 = 1.2816(0.8) + 3.5 = 4.52
Therefore 4.52 inches of precipitation must fall for it to be classified as extremely wet.
241/21
Suffolk University
STATS 250
Antonín Koukolík
A person must score in the upper 2% of the population on an IQ test to qualify for a membership in
Mensa, the international high-IQ society (U.S. Airways Attaché, September 2000). If IQ scores are
normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have
to qualify for Mensa? (1 point)
𝑥 = 𝑧𝜎 + 𝜇 = 2.0537(15) + 100 = 130.8
Therefore a person must score at least 130.8 in order to qualify for Mensa.
242/23
The time needed to complete a final examination in a particular college course is normally distributed
with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions.
a. What is the probability of completing the exam in one hour or less? (1 point)
𝑃(𝑥 < 60) = 𝑃 (𝑧 <
60 − 80
) = 𝑃(𝑧 < −2) = 0.02275
10
b. What is the probability that a student will complete the exam in more than 60 minutes but less
than 75 minutes? (1 point)
60 − 80
75 − 80
𝑃(60 < 𝑥 < 75) = 𝑃 (
<𝑧<
) = 𝑃(−2 < 𝑥 < −0.5) = 0.2857
10
10
c. Assume that the class has 60 students and that the examination period is 90 minutes in length.
How many students do you expect will be unable to complete the exam in the allotted time? (1
point)
90 − 80
) = 𝑃(𝑧 > 1) = 0.1587
10
Number of students who will be unable to complete: 0.1587(60) = 9.52 ≅ 10 students.
𝑃(𝑥 > 90) = 𝑃 (𝑧 >
276/18
Suffolk University
STATS 250
Antonín Koukolík
A population has a mean of 200 and a standard deviation of 50. A simple random sample of size 100 will
be taken and the sample mean 𝑥̅ will be used to estimate the population mean.
a. What is the expected value of 𝑥̅ ? (0.5 points)
𝐸(𝑥̅ ) = 𝜇𝑥 = 200
b. What is the standard deviation of 𝑥̅ ? (0.5 points)
𝜎𝑥̅ =
𝜎
√𝑛
=
50
=5
10
c. Show the sampling distribution of 𝑥̅ . (1 point)
𝑥̅ ~ 𝑁(200,5)
d. What does the sampling distribution of 𝑥̅ show? (1 point)
Probability distribution of 𝑥̅ .
277/19
A population has a mean of 200 and a standard deviation of 50. Suppose a simple random sample of size
100 is selected and 𝑥̅ is used to estimate μ.
a. What is the probability that the sample mean will be within ±5 of the population mean? (1 point)
𝑃(195 < 𝑥 < 205) = 𝑃 (
195 − 200
205 − 200
<𝑧<
) = 𝑃(−1 < 𝑧 < 1) = 0.6827
5
5
277/20
Suffolk University
STATS 250
Antonín Koukolík
Assume the population standard deviation is σ = 25.
a. Compute the standard error of the mean, 𝜎𝑥̅ , for sample sizes of 50. (0.5 points)
25
𝜎𝑥̅ =
= 3.536
√50
b. Compute the standard error of the mean, 𝜎𝑥̅ , for sample sizes of 100. (0.5 points)
𝜎𝑥̅ =
25
√100
= 2.5
c. Compute the standard error of the mean, 𝜎𝑥̅ , for sample sizes of 150. (0.5 points)
𝜎𝑥̅ =
25
√150
= 2.041
d. Compute the standard error of the mean, 𝜎𝑥̅ , for sample sizes of 200. (0.5 points)
𝜎𝑥̅ =
25
√200
= 1.768
e. What can you say about the size of the standard error of the mean as the sample size is
increased? (1 point)
As the sample size increases, the size of the standard error of the mean reduces.
Suffolk University