Assignment 7 (Chapter 6)
MA223 Summer 2025
Show work to get full credit!!
1. The normal distribution curve can be used as a probability distribution curve for
normally distributed variables.
A) True
B) False
2. When finding the area under the curve between two z values, the procedure is to
subtract the two z values and look up the corresponding area in Table E of Appendix C.
A) True
B) False
Look up both z values and then subtract the corresponding areas
3. The average hourly wage of workers at a fast food restaurant is $6.50/hr with a standard deviation of
$0.45. Assume that the distribution is normally distributed. If a worker at this fast food restaurant is
selected at random, what is the probability that the worker earns more than $6.75?
A) 28.77%
B) 27.64%
C) 42.07%
D) 5.17%
Mean = 6.50 | standard deviation = 0.45 | X = 6.75
P(X > 6.75)
Z = (x – mean) / sd
Z = (6.75 – 6.5) / 0.45 = 0.56 = z
Z = 0.56 = 0.7123
1 – 0.7123 = 0.2877 * 100 = 28.77%
4. The area under the curve is always positive even if the z value is negative.
A) True
B) False
5. What is the z value to the right of the mean such that 85% of the total area lies to the left of it as
shown in the figure below?
Z = 1.04
6. The finite population correction factor is necessary if relatively large samples are taken
from a small population, because the sample mean will then more accurately estimate
the
population mean and there will be less error in the estimation.
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7. A(n) sampling distribution of sample means is a distribution obtained by using the
means computed from random
samples of a specific size taken from a population.
8. In the figure below, what is the area under the curve between z = 1.50 and z = 2.50 ?
A) 0.0802
B) 0.0606
C) 0.0764
D) 1.00
Z2 – Z1 = area
Z(2.5) = 0.9938 | Z(1.5) = 0.9332
0.9938 – 0.9932 = 0.0606
9. A recent study found that the average life expectancy of a person living in Africa is 53
years with a standard deviation of 7.5 years. If a person in Africa is selected at random,
what is the probability that the person will die before the age of 65?
A) 94.52%
B) 82.89%
C) 94.95%
D) 88.49%
Mean = 53 | sd = 7.5 | x = 65
Z = (x – mean) / sd
Z = (65 – 53) / 7.5 = 1.6
Z = 1.6 = 0.9452 * 100 = 94.52%
10. The area under the normal distribution curve that lies within three standard deviations of
the mean is approximately 95%.
A) True
B) False – 2 standard deviations
11. The lengths of pregnancies in a small rural village are normally distributed with a mean of 265
days and a standard deviation of 15 days.
Mean = 265 | sd = 15
a. In what range would you expect to find the middle 50% of most pregnancies?
Z(0.25) = -0.67 | z(0.75) = 0.67
X = mean + z * sd
X(0.25) = 265 + (-0.67 * 15) = 254.95
X(0.75) = 265 + (0.67 * 15) = 275.05
IQR = 254.95 – 275.05 days
b. If you were to draw samples of size 36 from this population, in what range would you
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expect to find the middle 50% of most averages for the lengths of pregnancies in the
sample?
Mean = 265 | z1 = -0.67 | z2 = 0.67 | n = 36 | sd = 15
sd(sample means) = sd / sqrt(n)
sd(sample means) = 15 / sqrt(36) = 15 / 6 = 2.5
x = mean + (z * sample means)
x1 = 265 + (-0.67 * 2.5) = 263.325
x2 = 265 + (0.67 * 2.5) = 266.675
IQR = 263.325 – 266.675 days
12. A magazine reported that 6% of American drivers read the newspaper while driving. If
500 drivers are selected at random, find the probability that exactly 40 will admit to
reading the newspaper while driving.
A) 1.96%
B) 2.04%
C) 1.28%
D) 0.56%
p = 0.06 | n = 500 | q = 0.94 (1 – 0.06) | x = 40
mean = n * p | sd = sqrt(n * p * q)
mean = 500 * 0.06 = 30 | sd = sqrt(500 * 0.06 * 0.94) = 5.31
cont. correction use limits (39.5 < x < 40.5)
z = (x – mean) / sd
z1 = (39.5 – 30) / 5.31 = 1.79 = 0.9633
z2 = (40.5 – 30) / 5.31 = 1.98 = 0.9761
P(39.5 < x < 40.5) = 0.9761 – 0.9633 = 0.0128 * 100 = 1.28%
13. Find the area under the curve to the left of z = 1.69 .
A) 0.4545
B) 0.4452
C) 0.9545
D) 0.9452
Z = 1.69 = 0.9545
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14. Give the term for the number of standard deviations that a particular X value is away
from the mean.
A) z value
B) discrete value
C) continuous value
D) y value
15. Bolts that are used in the construction of an electric transformer are supposed to be
0.060 inches in diameter, and any bolt with diameter less than 0.058 inches or greater
than 0.062 inches must be scrapped. The machine that makes these bolts is set to
produce bolts of 0.060 inches in diameter, but it actually produces bolts with diameters
following a normal distribution with μ = 0.060 inches and σ= 0.001 inches. What is the
proportion of bolts that must be scrapped?
Mean = 0.060 | sd = 0.001 | 0.058 <= x <= 0.062
Z = (x – mean) / sd
Z1 = (0.058 – 0.060) / 0.001 = -2.00 = 0.0228
Z2 = (0.062 – 0.060) / 0.001 = 2.00 = 0.9772
P(-2 < Z < 2) = 0.9772 – 0.0228 = 0.9544
P(scrap) = 1 – 0.9544 = 0.0456
16. The 𝑝𝑟𝑜𝑏( 0 < 𝑋 < 2.25) is 0.4878.
A) True
0.9878 - .5 = 0.4878
B) False
17. What is the special property of the standard normal distribution?
A) The mean is 0 and the standard deviation is 1.
B) The total area under the normal distribution curve is equal to 1.00.
C) The curve is continuous.
D) The mean is located at the center of the distribution.
18. The marks on a statistics test are normally distributed with a mean of 62 and a variance of 225.
If the instructor wishes to assign B’s or higher to the top 30% of the students in the class, what
mark is required to get a B or higher?
Mean = 62 | sd = sqrt(225) = 15 | z = 1 – 0.3 = 0.7 (for top 30%)
Z = 0.7 = 0.52
X = mean + z * sd
X = 62 + (15 * .52)
X = 69.8
19. Stating that the area under the curve between z = 0 and z = 1.00 is 0.3413 is the same as
stating that the probability of selecting any z value between 0 and 1.00 is 0.3413.
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20. Using the normal distribution curve shown in the figure below, find the area under the
curve between z = 0 and z = −2.16 ?
A) –0.4821
B) 0.4821
C) –0.4846
D) 0.4846
Z = 0 = 0.5
Z = -2.16 = 0.0154
0.5 – 0.0154 = 0.4846
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