There are 505 full-service restaurants in North Dakota. The mean number of seats per restaurant is 74.1. [Source: Data based on the 2002 Economic Census from the US Census Bureau.] Suppose that the true population mean µ = 74.1 and standard deviation σ = 21 are unknown to the North Dakota tourism board. They select a simple random sample of 45 full-service restaurants located within the state to estimate µ. The mean number of seats per restaurant in the sample is M = 78.3, with a sample standard deviation of s = 17.8.
The standard deviation of the distribution of sample means (that is, the standard error, σM
M
) is . (Note: Although µ and σ are unknown to the North Dakota 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 505 full-service restaurants in North Dakota (µ = 74.1) and one randomly selected full-service restaurant in North Dakota is .
The standard or typical average difference between the mean number of seats in the sample of 45 restaurants (M = 78.3) and one randomly selected restaurant in that sample is .
The standard or typical average difference between the mean number of seats in the 505 full-service restaurants in North Dakota (µ = 74.1) and the sample mean of any sample of size 45 is .
The z-score that locates the mean number of seats in the North Dakota tourism board’s sample (M = 78.3) 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:
z
B: Proportion in Body
C: Proportion in Tail
D: 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 North Dakota tourism board selected a simple random sample of 45 full-service restaurants located within the state. Considering all possible such samples with n = 45, what is the probability of selecting one whose mean is greater than 78.3? That is, p(M > 78.3) = .