Question

Note: Please make sure to properly format your answers. All dollar figures in the answers need to include the dollar sign and any amount over 1,000 should include the comma ($2,354.67). All percentage values in the answers need to include a percentage sign (%). For all items without specific rounding instructions, round your answers to two decimal places, show both decimal places (5.06).
A recent survey by the American Automobile Association showed that a family of two adults and two children on vacation in the United States will pay an average of $247 per day for food and lodging with a standard deviation of $60 per day. If the data are normally distributed, find the percentage of these families who spent:
a. Less than $167 per day. _____

b. Less than $367 per day. _____

c. More than $247 per day. _____

d. More than $350 per day. _____

e. Less than $67 per day. ______

f. Between $200 and $300 per day. _____

g. Between $360 and $400 per day. _____

h. More than the median. ____

i. Less than the mean. _____

Answer 1

Using the normal distribution, the percentages are given as follows:

a) 9.18%.

b) 97.72%.

c) 50%.

d) 4.27%.

e) 0.13%.

f) 59.29%.

g) 2.46%.

h) 50%.

i) 50%.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

For this problem, the mean and the standard deviation are given as follows:

$\mu = 247, \sigma = 60$

For item a, the proportion is the p-value of Z when Z = 167, hence:

$Z = \frac{X - \mu}{\sigma}$

Z = (167 - 247)/60

Z = -1.33.

Z = -1.33 has a p-value of 0.0918.

Hence the percentage is of 9.18%.

For item b, the proportion is the p-value of Z when Z = 367, hence:

$Z = \frac{X - \mu}{\sigma}$

Z = (367 - 247)/60

Z = 2.

Z = 2 has a p-value of 0.9772.

Hence the percentage is of 97.72%.

For item c, the proportion is one subtracted by the p-value of Z when X = 247, hence:

$Z = \frac{X - \mu}{\sigma}$

Z = (247 - 247)/60

Z = 0

Z = 0 has a p-value of 0.5.

Hence the percentage is of 50%.

For item d, the proportion is one subtracted by the p-value of Z when X = 350, hence:

$Z = \frac{X - \mu}{\sigma}$

Z = (350 - 247)/60

Z = 1.72

Z = 1.72 has a p-value of 0.9573.

1 - 0.9573 = 0.0427.

Hence the percentage is of 4.27%.

For item e, the proportion is the p-value of Z when Z = 67, hence:

$Z = \frac{X - \mu}{\sigma}$

Z = (67 - 247)/60

Z = -3.

Z = -3 has a p-value of 0.0013.

Hence the percentage is of 0.13%.

For item f, the proportion is the p-value of Z when X = 300 subtracted by the p-value of Z when X = 200, hence:

X = 300:

$Z = \frac{X - \mu}{\sigma}$

Z = (300 - 247)/60

Z = 0.88.

Z = 0.88 has a p-value of 0.8106.

X = 200:

$Z = \frac{X - \mu}{\sigma}$

Z = (200 - 247)/60

Z = -0.78.

Z = -0.78 has a p-value of 0.2177.

0.8106 - 0.2177 = 0.5929.

Hence the percentage is 59.29%.

For item g, the proportion is the p-value of Z when X = 400 subtracted by the p-value of Z when X = 360, hence:

X = 400:

$Z = \frac{X - \mu}{\sigma}$

Z = (400 - 247)/60

Z = 2.55.

Z = 2.55 has a p-value of 0.9946.

X = 360:

$Z = \frac{X - \mu}{\sigma}$

Z = (360 - 247)/60

Z = 1.88.

Z = 1.88 has a p-value of 0.97.

0.9946 - 0.97 = 0.0246

Hence the percentage is 2.46%.

For items h and i, the distribution is symmetric, hence median = mean and the percentages are of 50%.

More can be learned about the normal distribution at brainly.com/question/24808124

#SPJ1