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There are 559 full-service restaurants in Delaware. The mean number of seats per restaurant is 99.2. [ Source : Data based on the 2002 Economic

There are 559 full-service restaurants in Delaware. The mean number of seats per restaurant is 99.2. [Source: Data based on the 2002 Economic Census from the US Census Bureau.]

Suppose that the true population mean = 99.2 and standard deviation = 21 are unknown to the Delaware tourism board. They select a simple random sample of 50 full-service restaurants located within the state to estimate . The mean number of seats per restaurant in the sample is M = 103.4, with a sample standard deviation of s = 18.2.

The standard deviation of the distribution of sample means (that is, the standard error,

M

M) is . (Note: Although and are unknown to the Delaware tourism board, they are known to you for the purposes of calculating these answers.)

The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware ( = 99.2) and one randomly selected full-service restaurant in Delaware is .

The standard or typical average difference between the mean number of seats in the sample of 50 restaurants (M = 103.4) andonerandomly selected restaurantin that sampleis .

The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (=99.2) and thesample meanof any sample of size 50 is .

The z-score that locates the mean number of seats in the Delaware tourism board's sample (M = 103.4) in the distribution of sample means is .

Use the unit normal tables and accompanying figures to answer the question that follows. To use the tables, select the desired range of z-score values. A table of the proportions of the normal distribution corresponding to that range of z-scores will appear.

Suggestion:Make a sketch of the area under the normal distribution you are seeking. This sketch will help you determine which column(s) of the normal table to use in determining the appropriate probability.

0.00 z 0.24:

0.25 z 0.49:

0.50 z 0.74:

0.75 z 0.99:

1.00 z 1.24:

1.25 z 1.49:

1.50 z 1.74:

1.75 z 1.99:

2.00 z 2.24:

2.25 z 2.49:

2.50 z 2.74:

2.75 z 2.99:

3.00 z 3.24:

3.30 z 4.00:

zB: Proportion in BodyC: Proportion in TailD: Proportion Between Mean and z.00.5000.5000.0000.01.5040.4960.0040.02.5080.4920.0080.03.5120.4880.0120.04.5160.4840.0160.05.5199.4801.0199.06.5239.4761.0239.07.5279.4721.0279.08.5319.4681.0319.09.5359.4641.0359.10.5398.4602.0398.11.5438.4562.0438.12.5478.4522.0478.13.5517.4483.0517.14.5557.4443.0557.15.5596.4404.0596.16.5636.4364.0636.17.5675.4325.0675.18.5714.4286.0714.19.5753.4247.0753.20.5793.4207.0793.21.5832.4168.0832.22.5871.4129.0871.23.5910.4090.0910.24.5948.4052.0948

The Delaware tourism board selected a simple random sample of 50 full-service restaurants located within the state. Considering all possible such samples with n = 50, what is the probability of selecting one whose mean is greater than 103.4? That is, p(M > 103.4) = .

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