Question

In: Math

In a recent year the average movie ticket cost $10.50, In a random sample of 50...

In a recent year the average movie ticket cost $10.50, In a random sample of 50 movie tickets from various areas

What is the probability that the mean cost exceeds $8.50, given that the population standard deviation is $1.50?

Expert Solution

Solution :

mean = = 10.50

standard deviation = = 1.50

n = 50

= = 10.50

= / n = 1.50/ 50 = 0.21

P( > 8.50) = 1 - P( < 8.50)

= 1 - P[( - ) / < (8.50 - 10.50) / 0.21]

= 1 - P(z < -9.52)

Using z table,

= 1 - 0

= 1

milcah answered 9 months ago

A company has 189 accountants. In a random sample of 50 of them, the average number...

A company has 189 accountants. In a random sample of 50 of them, the average number of overtime hours worked in a week was 9.7 and the sample standard deviation was 6.2 hours. a) Find a 95% confidence interval of the average number of overtime hours worked by each accountant in this company during that week. b) Find a 99% confidence interval of the total number of overtime hours worked by each accountant in this company during that week.

In a random sample of 50 refrigerators, the mean repair cost was $132.00 and the population...

In a random sample of 50 refrigerators, the mean repair cost was $132.00 and the population standard deviation is $15.10 A 95% confidence interval for the population mean repair cost is (127.81,136.19). Change the sample size to n=100 Construct a 95 confidence interval for the population mean repair cost. Which confidence interval is wider? Explain.

4. When the ticket price for a concert at the opera house was $50, the average...

4. When the ticket price for a concert at the opera house was $50, the average attendance was 4000 people. When the ticket price was raised to $52, the average attendance was 3800 people. a. Assuming the demand function is linear, find the demand function, p. b. Find the number of tickets sold that maximize the revenue. Use the second derivative test to verify it is a maximum. c. Find the price that maximizes the revenue. d. Find the maximum...

A recent survey shows that the average of American household’s annual expenditure on coffee is $54.16 with a standard deviation of $10.50.

A recent survey shows that the average of American household’s annual expenditure on coffee is $54.16 with a standard deviation of $10.50. We randomly select a household. Assuming that the expenditure follows normal distribution, find: a)The probability that their annual expenditure on coffee is less than $25? b)The probability that it is more than $50? c)The probability that it is between $40 and $60? d)The amount less than which 95% of households spend on coffee?

A random sample of 20 recent weddings in a country yielded a mean wedding cost of...

A random sample of 20 recent weddings in a country yielded a mean wedding cost of $26,327.16 Assume that recent wedding costs in this country are normally distributed with a standard deviation of $8,100 Complete parts (a) through (c) below. a. Determine a 95% confidence interval for the mean cost, muμ, of all recent weddings in this country. The 95% confidence interval is from ????? (Round to the nearest cent as needed.)

In a random sample of 50 Americans five years ago (Group 1), the average credit card...

In a random sample of 50 Americans five years ago (Group 1), the average credit card debt was $5,779. In a random sample of 50 Americans in the present day (Group 2), the average credit card debt is $6,499, Let the population standard deviation be $1,152 five years ago, and let the current population standard deviation be $1,634. Using a 0.01 level of significance, test if there is a difference in credit card debt today versus five years ago. What...

A random sample of 50 measurements resulted in a sample mean of 62 with a sample...

A random sample of 50 measurements resulted in a sample mean of 62 with a sample standard deviation 8. It is claimed that the true population mean is at least 64. a) Is there sufficient evidence to refute the claim at the 2% level of signifigance? b) What is the p-vaule? c) What is the smallest value of alpha for which the claim will be rejected?

A random sample of 800 movie goers in Flagstaff found 328 movie goers who had bought...

A random sample of 800 movie goers in Flagstaff found 328 movie goers who had bought popcorn on their last visit. Find a 95% confidence interval for the true percent of movie goers in Flagstaff who have bought popcorn on their last visit. Express your results to the nearest hundredth of a percent.

From a random sample of 48 days in a recent year, US Gasoline prices had a...

From a random sample of 48 days in a recent year, US Gasoline prices had a mean of $2.34 and a standard deviation of $0.30. Use this information to construct the following confidence intervals and determine the margin of error for each: a. 90% b. 95% c. 99%

Using a random sample of n = 50, the sample mean is = 13.5. Suppose that...

Using a random sample of n = 50, the sample mean is = 13.5. Suppose that the population standard deviation is σ=2.5. Is the above statistical evidence sufficient to make the following claim μ ≠15: ?o: μ=15 ??: μ ≠15 α = 0.05. p value = 0 Interpret the results using the p value test. Reject Ho or Do not reject Ho