In: Statistics and Probability
1. At a certain restaurant in Ohio, the number of minutes that diners spend at the table has a major impact on the profitability of the restaurant. Suppose the average number of minutes that diners spend at a table for dinner at the restaurant is 97 minutes with a standard deviation of 18 minutes. Assume the number of minutes diners spend at their table follows the normal probability distribution. Complete parts a through d.
a. Calculate the probability that the average number of minutes that diners spend at their table for dinner will be less than 100 minutes using a sample size of 9 tables.
(Round to three decimal places as needed.)
2. According to the Bureau of Labor Statistics, Americans spent on average $2, 913 in 2016 on entertainment. Assume the population standard deviation of this spending is $863. A random sample of 28 adults was selected and was found to have an average spending of $2,725 on entertainment. Complete parts a and b.
a. Does this sample provide support for the conclusions of the BLS poll?
There is a .... chance of observing a sample mean as low as $2 comma 7252,725. This probability is ▼ low not low enough (0.05) to contradict the findings of the poll.
(Type an integer or decimal rounded to three decimal places as needed.)
3. The average weight of a professional football player in 2009 was 246.2 pounds. Assume the population standard deviation is 25pounds. A random sample of 32 professional football players was selected. Complete parts a through e.
a. Calculate the standard error of the mean. σ-x
4. According to a certain organization, adults worked an average of 1,783 hours last year. Assume the population standard deviation is 370 hours and that a random sample of 50 adults was selected. Complete parts a through e below.
a. Calculate the standard error of the mean. sigma Subscript x overbarσxequals=n
(Round to two decimal places as needed.)
5. A college has 250 full-time employees that are currently covered under the school's health care plan. The average out-of-pocket cost for the employees on the plan is $11,910 with a standard deviation of $520. The college is performing an audit of its health care plan and has randomly selected 35 employees to analyze their out-of-pocket costs.
a. Calculate the standard error of the mean.
b. What is the probability that the sample mean will be less than $1,860?
c. What is the probability that the sample mean will be more than $1,880?
d. What is the probability that the sample mean will be between $1,930 and $1,960?
a. The standard error of the mean is nothing.
(Round to two decimal places as needed.)
1.
Standard error of mean = =
18 / = 6
Probability that average number of minutes that diners spend at their table for dinner will be less than 100 minutes
= P( < 100)
= P[Z < (100 - 97)/6]
= P[Z < 0.5]
= 0.691
2.
= 2913
= 863
Standard error of mean = = 863 / = 163.0917
Probability that average spending is lower than $2,725 = P[ < 2725]
= P[Z < (2725 - 2913) / 163.0917]
= P[Z < -1.15]
= 0.1251
There is a 0.1251 chance of observing a sample mean as low as $2,725. This probability is not low enough (0.05) to contradict the findings of the poll.
3.
Standard error of mean = = 25 / = 4.419417
4.
Standard error of mean = = 370 / = 52.33
5.
a.
For finite population, Standard error of mean =
= 81.67
b.
Probability that the sample mean will be less than $1,860 = P[ < 1860]
= P[Z < (1860 - 1910)/81.67]
= P[Z < -0.61]
= 0.2709
c.
Probability that the sample mean will be more than $1,880 = P[ < 1880]
= P[Z > (1880 - 1910)/81.67]
= P[Z > -0.37]
= 0.6443
d.
Probability that the sample mean will be between$1,930 and $1,960 = P[1930 < < 1960]
=P[ < 1960] - P[ < 1930]
= P[Z < (1960 - 1910)/81.67] - P[Z < (1930 - 1910)/81.67]
= P[Z < 0.61] - P[Z < 0.24]
= 0.7291 - 0.5948
= 0.1343