A small university knows the average amount that its students spend on lunch each day. The...

A small university knows the average amount that its students spend on lunch each day. The amount spent on lunches for the population of 500 students is not highly skewed and has a mean of $8 and a standard deviation of $2. Suppose simple random sample of 49 students is taken, what is the probability that the sample mean for the sample of 49 students will be between $7.50 and $8.50?

Solutions

Expert Solution

Solution :

Given that ,

mean = = 8

standard deviation = = 2

n = 49

= 8

= / $\sqrt$ n= 2/ $\sqrt$ 49 =0.29

P(7.50< <8.50 ) = P[(7.50-8) / 0.29< ( - ) / < (8.50-8) /0.29 )]

= P( -1.72< Z <1.72 )

= P(Z <1.72 ) - P(Z <-1.72 )

Using z table

=0.9573-0.0427

=0.9146

probability= 0.9146

orchestra answered 11 months ago

Next > < Previous

Related Solutions

Students of a large university spend an average of $5 a day on lunch. The standard...

Students of a large university spend an average of $5 a day on lunch. The standard deviation of the expenditure is $3. A simple random sample of 36 students is taken. What are the expected value, standard deviation, and shape of the sampling distribution of the sample mean? What is the probability that the sample mean will be at least $4? What is the probability that the sample mean will be at least $5.90?

Students of a large university spend an average of $6 a day on lunch. The standard...

Students of a large university spend an average of $6 a day on lunch. The standard deviation of the expenditure is $2. A simple random sample of 81 students is taken. 1. What is the probability that the sample mean will be at least $5.25? 2. What is the probability that the sample mean will be at least $6.50? 3. What is the range of money spent by people who fall within one standard deviation of the mean? 4. Kelsey...

Students of a large university spend an average of $7 a day on lunch. The standard...

Students of a large university spend an average of $7 a day on lunch. The standard deviation of the expenditure is $2. A simple random sample of 25 students is taken. What is the probability that the sample mean will be at least $4? Jason spent $15 on his lunch. Explain, in terms of standard deviation, why his expenditure is not usual. Explain what information is given on a z table. For example, if a student calculated a z value...

The average amount you spend on a lunch during the week is not known. Based on...

The average amount you spend on a lunch during the week is not known. Based on past experience, you are willing to assume that the standard deviation is $2.10. You take a random sample of 28 lunches, and apply the central limit theorem. Round your answer to 2 decimal places. Fill in the blank: The 68–95–99.7 rule says that the probability is about 0.95 that is within $ ____ of the population mean mu.

A small diner is open for lunch each day. The lunch counter seats a maximum of...

A small diner is open for lunch each day. The lunch counter seats a maximum of 12 customers at one time. The average time required for a customer to complete lunch 30 minutes. It takes 4 minutes of a server’s time to take care of a customer. There are two servers assigned to the counter. The cook in the kitchen can prepare an order in 1.5 minutes. Answer the following questions, assuming the diner is relatively full and operating (i.e.,...

A survey of college students reported that they spend an average of $9.50 a day on...

A survey of college students reported that they spend an average of $9.50 a day on dinner with a standard deviation of $3. What is the probability that 100 randomly selected college students will spend less than $10.00 on average for dinner? Round your answer to 4 decimal places.

1. In order to test the claim that average amount spent for lunch per day by...

1. In order to test the claim that average amount spent for lunch per day by MSU students is not $ 7.25. A random sample of 40 students is taken. After calculating sample mean and standard deviation p- value is found to be 0.0482. Which one of the following statement is correct. Question 6 options: Reject H0 at 95% confidence level Reject H0 at 99% confidence level Reject H0 at 98% confidence level Reject H0 at 96% confidence level 2....

Applying the 68-95-99.7 rule. The average amount you spend on a lunch during the week is...

Applying the 68-95-99.7 rule. The average amount you spend on a lunch during the week is not known. Based on past experience, you are willing to assume that the standard deviation is $2.10. You are taking a random sample of 28 lunches. The 68-95-99.7 rule says that the probability is about 0.95 that x with bar on top is within $_________ of the population mean μ. Fill in the blank (report to two decimal places)

Is the average amount of time spent sleeping each day different between male and female students?...

Is the average amount of time spent sleeping each day different between male and female students? Which hypothesis test is the most appropriate to use in this problem? Why? (Note: if you are doing a 2-sample t-test, make sure you state which one you are doing and why.) If the test you chose has a t-statistic, report it here with degrees of freedom. If it does not, state that the test you chose does not have a test-statistic. Give and...

Data for the amount of time that Americans spend commuting to work each day creates a...

Data for the amount of time that Americans spend commuting to work each day creates a histogram that has a normal shape. The mean is 35 minutes and the standard deviation is 10 minutes. According to the Empirical Rule, which of the following is not a conclusion that we can make? a. It would be common to find a commuter who takes 43 minutes to commute to work. b. About 5% of commuters take longer than 55 minutes to commute...

Subjects