Continuous Random Variables & The Normal Dist’n
Name _______________________________
Read each problem carefully. Write your answer in the blank, or circle the correct answer.
1. A study was conducted that resulted in the following relative frequency histogram. Determine
whether or not the histogram indicates that a normal distribution could be used as a model for the
variable. a) The histogram is not bell-shaped, so a normal
distribution could not be used as a model. b) The histogram is bell-shaped, so a normal distribution
could be used as a model. c) Not enough information
2. One graph in the figure represents a normal distribution with µ = 6 and σ = 2.
The other graph represents a normal distribution with µ = 15 and σ = 2.
Determine which graph is which and explain how you know. a) Graph A has µ = 15 and Graph B has µ = 6 because a larger mean shifts the graph to the right. b) Graph A has µ = 15 and Graph B has µ = 6 because a larger mean shifts the graph to the left. c) Graph A has µ = 6 and Graph B has µ = 15 because a larger mean shifts the graph to the right. d) Graph A has µ = 6 and Graph B has µ = 15 because a larger mean shifts the graph to the left.
3. The graph of a normal curve is given on the right. Use the graph to identify the values of µ and σ.
µ = __________
σ = __________

a)
4. Suppose the monthly charge for cell phone plans is normally distributed with µ = $56 and σ = $15.
Which of the following is the correct normal curve with the correct parameters labeled? b) c)
Which of the following shows the shaded region that represents the proportion of plans that charge less than
$41? a) b) c)
Suppose the area under the normal curve to the left of $41 is 0.1587. Which of the following is the correct interpretation of this result? a) 84.13% of cell phone plans cost less than $41 a month. b) 15.87% of cell phone plans cost less than $41 a month. c) The probability is 0.1587 that a randomly selected cell phone plan is more than $41 a month. d) 15.87% of cell phone plans cost more than $41 a month.
5. Determine the area under the standard normal curve that lies to the left of Z = 0.43
(round to 4 decimals)
Area = _______________
6. Calculate P(Z < –1.22).
(round to 4 decimals) P(Z < –1.22) = ______________
7. Determine the area under the standard normal curve that lies to the right of Z = –1.38.
(round to 4 decimals)
Area = _______________
8. Calculate P(Z > 0.24).
(round to 4 decimals) P(Z > 0.24) = ______________

9. Determine the area under the standard normal curve that lies between Z = –1.94 and Z = 1.94.
(round to 4 decimals)
Area = _______________
10. Calculate P(0.68 < Z < 2.04).
P(0.68 < Z < 2.04) = ______________ (round to 4 decimals)
11. Find the z-score such that the area under the standard normal curve to the left is 0.36.
(round to 2 decimals)
Z = _______________
12. Find the z-score such that the area under the standard normal curve to the right is 0.37.
(round to 2 decimals)
Z = _______________
13. Find the z-scores that separate the middle 48% of the distribution from the area in the tails of the
standard normal distribution.
(round to 2 decimals)
Z = _______________
14. Assume the random variable X is normally distributed with µ = 50 and σ = 7. Find the 67 th percentile.
(round to 2 decimals)
P
67
= _________________

15. The mean incubation time for a type of fertilized egg kept at 100.1°F is 21 days with a standard deviation
of 2 days. Suppose that the incubation times are approximately normally distributed.
What is the probability that a randomly selected fertilized egg hatches in less than 19 days?
(round to 4 decimals)
P(less than 19) = _____________
What is the probability that a randomly selected fertilized egg takes more than 23 days to hatch?
(round to 4 decimals)
P(more than 23) = _____________
What is the probability that a randomly selected fertilized egg hatches between 17 and 21 days?
(round to 4 decimals)
P(between 17 and 21) = ___________
Would it be unusual for an egg to hatch in less than 18 days? Why?
The probability of an egg hatching in less than 18 days is _______________, so it ________________
(4 decimals) would/would not
Be unusual because the probability is _________________ than 0.05 (5%).
greater/less
Determine the 20 th percentile for incubation times.
(round to a whole number)
P
20
= _______________days
Determine the cutoff incubation times that make up the middle 95% of fertilized eggs.
(round to a whole number)
Incubation times: ________ and _______ days

16. Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed
with µ = 132 days and σ = 11 days.
What is the probability that a randomly selected pregnancy last less than 128 days?
(round to 4 decimals)
P( x < 128) = _________________
What is the probability that a random sample of 13 pregnancies has a mean gestation period of less than 128 days? (round to 4 decimals)
P( x < 128) = _________________
What is the probability that a random sample of 41 pregnancies has a mean gestation period of less than 128 days? (round to 4 decimals)
P( x < 128) = _________________
17. The reading speed of second grade students is approximately normal, with a mean of 89 words per minute
and a standard deviation of 10 words per minute.
What is the probability that a randomly selected student will read more than 95 words per minute?
(round to 4 decimals)
P( x > 95) = _________________
What is the probability that a random sample of 10 second grade students results in a mean reading rate of more than 95 words per minute? (round to 4 decimals)
P( x > 95) = _________________
What is the probability that a random sample of 20 second grade students results in a mean reading rate of more than 95 words per minute? (round to 4 decimals)
P( x > 95) = _________________