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dolphi86 [110]
3 years ago
15

A random sample of 128 lunch customers was selected at a restaurant. The average amount of time the customers in the sample stay

ed in the restaurant was 128.2 minutes and the sample standard deviation equals 24.9 minutes. The objective is to create a 99% confidence interval for the mean amount of time that customers stay. What is the confidence interval?
Mathematics
1 answer:
matrenka [14]3 years ago
3 0

Answer: (122.52, 133.88)

Step-by-step explanation:

We are to construct a 99% confidence interval for population mean time.

From our question, we have the following parameters

Sample size (n) = 128

Sample mean (x) = 128.2

Sample standard deviation (s) = 24.9

To construct a 99% confidence interval, we use the formulae below.

u = x + Zα/2 × s/√n........ For upper limit

u = x - Zα/2 × s/√n........ For lower limit

We are using a z score for our critical value (Zα/2) and that's because our sample size is greater than 30 (n = 128), even though we have our sample standard deviation.

The value of Zα/2 is gotten using a z distribution table and has a value of 2.58

For lower limit, we have that

u = 128.2 - 2.58/ (24.9/√128)

u = 128.2 - 2.58 (2.2)

u = 128.2 - 5.678

u = 122.52.

For upper limit, we have that

u = 128.2 + 2.58/ (24.9/√128)

u = 128.2 + 2.58 (2.2)

u = 128.2 + 5.678

u = 133.88

Hence the 99% confidence interval for the mean amount of time customer stays is (122.52, 133.88)

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Step-by-step explanation:

An elf ate 15 of your muffins and that was 5/8 of all of them. To get the number of muffins left goes thus:

We can first calculate the total number of muffins the person had. Let the total number of muffins be y. That means the elf ate 5/8 of y.

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