Question

In: Statistics and Probability

1. At a certain restaurant in​ Ohio, the number of minutes that diners spend at the...

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.)

Solutions

Expert Solution

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


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