Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of…
- Isle Royale, an island in Lake Superior, has provided an important study site of wolves and their prey. Of special interest is the study of the number of moose killed by wolves. In the period from 1958 to 1974, there were 296 moose deaths identified as wolf kills. The age distribution of the kills is as follows.
| Age of Moose in Years | Number Killed by Wolves |
| Calf (0.5 yr)
1-5 6-10 11-15 16-20 |
106
50 70 67 3 |
(a) For each age group, compute the probability that a moose in that age group is killed by a wolf. (Use 3 decimal places.)
| 0.5 | |
| 1-5 | |
| 6-10 | |
| 11-15 | |
| 16-20 |
(b) Consider all ages in a class equal to the class midpoint. Find the expected age of a moose killed by a wolf and the standard deviation of the ages. (Use 2 decimal places.)
| μ | |
| σ |
- According to Harper’s Index, 50%of all federal inmates are serving time for drug dealing. A random sample of 15federal inmates is selected.
(a) What is the probability that 12 or more are serving time for drug dealing? (Use 3 decimal places.)
(b) What is the probability that 3 or fewer are serving time for drug dealing? (Use 3 decimal places.)
(c) What is the expected number of inmates serving time for drug dealing? (Use 1 decimal place.)
- The Orchard Cafe has found that about 4%of the diners who make reservations don’t show up. If 82reservations have been made, how many diners can be expected to show up? Find the standard deviation of this distribution. (Use 2 decimal places.)
| μ | |
| σ |
- Look at the normal curve below, and find μ, μ+ σ, and σ.
| μ | = | |
| μ + σ | = | |
| σ | = |
- What percentage of the area under the normal curve lies as given below?
(a) to the right of μ
%
(b) between μ – 2σ and μ + 2σ
%
(c) to the right of μ + 3σ (Use 2 decimal places.)
%
- The incubation time for Rhode Island Red chicks is normally distributed with a mean of 26days and standard deviation of approximately 3days. Look at the figure below and answer the following questions. If 1000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time. Assume all eggs eventually hatch.)
(a) in 20 to 32 days
chicks
(b) in 23 to 29 days
chicks
(c) in 26 days or fewer
chicks
(d) in 17 to 35 days
chicks
- Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean μ= 25.1kilograms and standard deviation σ = 3.3 kilograms. Let x be the weight of a fawn in kilograms.
Convert the following x intervals to z intervals. (Round your answers to two decimal places.)
(a) x < 30
z <
(b) 19 < x
< z
(c) 32 < x < 35
< z <
Convert the following z intervals to x intervals. (Round your answers to one decimal place.)
(d) −2.17 < z
< x
(e) z < 1.28
x <
(f) −1.99 < z < 1.44
< x <
(g) If a fawn weighs 14 kilograms, would you say it is an unusually small animal? Explain using z values and the figure above.
Yes. This weight is 3.36 standard deviations below the mean; 14 kg is an unusually low weight for a fawn.Yes. This weight is 1.68 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. No. This weight is 3.36 standard deviations below the mean; 14 kg is a normal weight for a fawn.No. This weight is 3.36 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.No. This weight is 1.68 standard deviations above the mean; 14 kg is an unusually high weight for a fawn.
(h) If a fawn is unusually large, would you say that the z value for the weight of the fawn will be close to 0, −2, or 3? Explain.
It would have a large positive z, such as 3.It would have a negative z, such as −2. It would have a z of 0.
- Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.)
The area to the left of
z = −1.33
is .
- Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.)
The area to the right of z = 1.55 is .
- Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.)
The area between z = 0 and z = 2.73 is .
- Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.)
The area between
z = −2.25
and z = 1.42 is .
- Let zbe a random variable with a standard normal distribution. Find the indicated probability. (Round your answer to four decimal places.)
P(−1.21 ≤ z ≤ 2.42)
=
Shade the corresponding area under the standard normal curve.
- Consider a normal distribution with mean 31and standard deviation 5. What is the probability a value selected at random from this distribution is greater than 31? (Round your answer to two decimal places.)
- Consider the following data. The summary statistics, histogram, and normal quantile plot were generated by Minitab.
| 27 | 27 | 27 | 28 | 28 | 28 | 28 | 28 | 28 | 29 |
| 29 | 29 | 29 | 29 | 29 | 29 | 29 | 29 | 29 | 30 |
| 30 | 30 | 30 | 30 | 30 | 30 | 30 | 30 | 30 | 30 |
| 30 | 31 | 31 | 31 | 31 | 31 | 31 | 31 | 31 | 32 |
| 32 | 32 | 32 | 33 | 33 | 33 | 33 | 33 | 34 | 34 |
| Variable
Data |
N
50 |
N*
0 |
Mean
30.160 |
SE Mean
0.256 |
StDev
1.811 |
| Variable
Data |
Minimum
27.000 |
Q1
29.000 |
Median
30.000 |
Q3
31.000 |
Maximum
34.000 |
(a) Does the histogram indicate normality for the data distribution? Explain.
No, we observe a bell-shaped, skewed distribution.No, we observe a uniform, skewed distribution. Yes, we observe a uniform, symmetric distribution.Yes, we observe a bell-shaped, symmetric distribution.
(b) Does the normal quantile plot indicate normality for the data distribution? Explain.
Yes, the points do not fall on a straight line.No, the points fall approximately on a straight line. Yes, the points fall approximately on a straight line.No, the points do not fall on a straight line.
(c) Compute the interquartile range.
Check for outliers.
There are two outliers.There are no outliers. There are no outliers, but two data points fall exactly on the boundary.There is one outlier.
(d) Compute Pearson’s index. (Round your answer to three decimal places.)
Does the index value indicate skewness?
YesNo
(e) Using parts (a) through (d), would you say the data are from a normal distribution?
No, the graphs do not support normality, there are no apparent outliers, but there is skewness.Yes, the graphs support normality, there are no apparent outliers, and there is no skewness. No, the graphs do not support normality, there is no skewness, but there are apparent outliers.Yes, the graphs support normality, there are no apparent outliers, but there is skewness.
- Assume that xhas a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)
μ = 4.1; σ = 2.5
P(3 ≤ x ≤ 6) =
- Assume that xhas a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)
μ = 5.7; σ = 0.8
P(7 ≤ x ≤ 9) =
- Assume that xhas a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)
μ = 25; σ = 3.5
P(x ≥ 30) =
- Find zsuch that 6.8%of the standard normal curve lies to the left of z. (Round your answer to two decimal places.)
z =
Sketch the area described.
- Quick Start Company makes 12-volt car batteries. After many years of product testing, the company knows that the average life of a Quick Start battery is normally distributed, with a mean of 45.4months and a standard deviation of 9.3months.
(a) If Quick Start guarantees a full refund on any battery that fails within the 36-month period after purchase, what percentage of its batteries will the company expect to replace? (Round your answer to two decimal places.)
%
(b) If Quick Start does not want to make refunds for more than 10% of its batteries under the full-refund guarantee policy, for how long should the company guarantee the batteries (to the nearest month)?
months
- Suppose xhas a distribution with μ= 42 and σ = 7.
(a) If random samples of size n = 16 are selected, can we say anything about the x distribution of sample means?
No, the sample size is too small.Yes, the x distribution is normal with mean μ x = 42 and σ x = 1.75. Yes, the x distribution is normal with mean μ x = 42 and σ x = 0.4.Yes, the x distribution is normal with mean μ x = 42 and σ x = 7.
(b) If the original x distribution is normal, can we say anything about the x distribution of random samples of size 16?
Yes, the x distribution is normal with mean μ x = 42 and σ x = 7.Yes, the x distribution is normal with mean μ x = 42 and σ x = 1.75. No, the sample size is too small.Yes, the x distribution is normal with mean μ x = 42 and σ x = 0.4.
Find P(38 ≤ x ≤ 43). (Round your answer to four decimal places.)
- Suppose xhas a distribution with μ= 15 and σ = 14.
(a) If a random sample of size n = 35 is drawn, find μx, σ x and P(15 ≤ x ≤ 17). (Round σx to two decimal places and the probability to four decimal places.)
| μx = |
| σ x = |
| P(15 ≤ x ≤ 17) = |
(b) If a random sample of size n = 74 is drawn, find μx, σ x and P(15 ≤ x ≤ 17). (Round σ x to two decimal places and the probability to four decimal places.)
| μx = |
| σ x = |
| P(15 ≤ x ≤ 17) = |
(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is
part (a) because of the sample size. Therefore, the distribution about μx is
.
- Let xbe a random variable that represents the weights in kilograms (kg) of healthy adult female deer (does) in December in a national park. Then xhas a distribution that is approximately normal with mean μ = 67.0 kg and standard deviationσ = 7.5 kg. Suppose a doe that weighs less than 58 kg is considered undernourished.
(a) What is the probability that a single doe captured (weighed and released) at random in December is undernourished? (Round your answer to four decimal places.)
(b) If the park has about 2750 does, what number do you expect to be undernourished in December? (Round your answer to the nearest whole number.)
does
(c) To estimate the health of the December doe population, park rangers use the rule that the average weight of n = 45 does should be more than 64 kg. If the average weight is less than 64 kg, it is thought that the entire population of does might be undernourished. What is the probability that the average weight
x
for a random sample of 45 does is less than 64 kg (assuming a healthy population)? (Round your answer to four decimal places.)
(d) Compute the probability that
x
< 69 kg for 45 does (assume a healthy population). (Round your answer to four decimal places.)
Suppose park rangers captured, weighed, and released 45 does in December, and the average weight was
x
= 69 kg. Do you think the doe population is undernourished or not? Explain.
Since the sample average is above the mean, it is quite likely that the doe population is undernourished.Since the sample average is below the mean, it is quite unlikely that the doe population is undernourished. Since the sample average is below the mean, it is quite likely that the doe population is undernourished.Since the sample average is above the mean, it is quite unlikely that the doe population is undernourished.
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