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 μ = 99.2 and standard deviation = 20.8 are unknown to the Delaware tourism board. They select a simple random sample of 50 full-service restaurants located within the state to estimate u. 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, GM) is 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 (μ = 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) and one randomly selected restaurant in that sample is The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (p = 99.2) and the sample mean of 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. . (Note: Although μ and σ are unknown to 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. Body B 0 Z Tail C 0 Z D: Between 0 and z
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 μ = 99.2 and standard deviation = 20.8 are unknown to the Delaware tourism board. They select a simple random sample of 50 full-service restaurants located within the state to estimate u. 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, GM) is 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 (μ = 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) and one randomly selected restaurant in that sample is The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (p = 99.2) and the sample mean of 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. . (Note: Although μ and σ are unknown to 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. Body B 0 Z Tail C 0 Z D: Between 0 and z
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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![5. Three different distributions
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 μ = 99.2 and standard deviation o = 20.8 are unknown to the Delaware tourism board. They select a
simple random sample of 50 full-service restaurants located within the state to estimate u. 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, OM) is
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 (u = 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) and one randomly
selected restaurant in that sample is
The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (u = 99.2) and the
sample mean of 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.
(Note: Although μ and o are unknown to
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.
Body B
0
Tail C
0
Z
-D: Between 0
and z](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf8c9471-3d77-40bb-8dfb-967f2c9588b8%2F5cc1c25a-21c8-488a-a98b-42096ccf6d6a%2Feyyole_processed.png&w=3840&q=75)
Transcribed Image Text:5. Three different distributions
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 μ = 99.2 and standard deviation o = 20.8 are unknown to the Delaware tourism board. They select a
simple random sample of 50 full-service restaurants located within the state to estimate u. 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, OM) is
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 (u = 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) and one randomly
selected restaurant in that sample is
The standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (u = 99.2) and the
sample mean of 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.
(Note: Although μ and o are unknown to
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.
Body B
0
Tail C
0
Z
-D: Between 0
and z
Transcribed Image Text:0.00 ≤z≤ 0.24:
2.00 ≤z≤ 2.24:
Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.10
.11
.12
.13
.14
.15
.16
.17
.18
.19
.20
.21
.22
.23
.24
0.25 sz≤ 0.49:
2.25 sz≤ 2.49:
0.50 ≤z≤ 0.74:
2.50 ≤z≤ 2.74:
.5000
.5040
.5080
.5120
.5160
.5199
.5239
.5279
.5319
.5359
.5398
.5438
.5478
.5517
.5557
.5596
.5636
.5675
.5714
.5753
.5793
.5832
.5871
.5910
.5948
0.75 ≤ z≤ 0.99:
2.75 sz≤ 2.99:
.5000
.4960
.4920
.4880
.4840
.4801
.4761
.4721
.4681
.4641
.4602
.4562
.4522
.4483
.4443
.4404
.4364
.4325
.4286
.4247
.4207
.4168
.4129
.4090
.4052
1.00 sz≤ 1.24:
3.00 ≤z≤ 3.24:
B: Proportion in Body C: Proportion in Tail D: Proportion Between Mean and z
.0000
.0040
.0080
.0120
.0160
.0199
.0239
.0279
.0319
.0359
.0398
.0438
.0478
.0517
.0557
.0596
.0636
.0675
.0714
.0753
.0793
.0832
.0871
.0910
.0948
1.25 ≤ z≤ 1.49:
3.30 ≤z≤ 4.00:
1.50 sz≤ 1.74:
1.75 sz≤ 1.99:
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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