Please help with my stats hw 6. Three different distributionsThere are 559 full-service restaurants in Delaware. The mean number of seats per restaurant is 99.2. [Source: Data based on the 2002 EconomicCensus from the US Census Bureau.]Suppose that the true population mean u = 99.2 and standard deviation o = 20.5 are unknown to the Delaware tourism board. They select a simplerandom 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(Note: Although p and o are unknown tothe 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 (p = 99.2) and onerandomly selected full-service restaurant in Delaware isThe standard or typical average difference between the mean number of seats in the sample of 50 restaurants (M = 103.4) and one randomlyselected restaurant in that sample isThe standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (u = 99.2) and thesample mean of any sample of size 50 isThe 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 isUse 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 thenormal table to use in determining the appropriate probability.D: Between 0and zBody BTail C 1.00 szs 1.24:1.25 szs 1.49:1.50 szs 1.74:1.75 szs 1.990.00 szs 0.24:0.50 szs 0.74:0.75 szs0.99:0.25 szs0.49:2.75 szs 2.99:3.00 szs 3.24:3.30 szS 4.00:2.00 szs 2.24:2.25 szs 2.49:2.50 szs 2.74:B: Proportion in BodyC: Proportion in TailD: Proportion Between Mean and z.5000.0000.00.5000.01.5040.4960.0040.02.5080.4920.0080.03.5120.4880.0120.04.5160.4840.0160.05.5199.4801.0199.06.5239.4761.0239.07.5279.4721.0279.08.5319.4681.0319.09.5359.4641.0359.10.5398.4602.0398.11.5438.4562.0438.12.5478.4522.0478.13.5517.4483.0517.14.5557.4443.0557.15.5596.4404.0596.16.5636.4364.0636.17.56754325.0675.18.5714.4286.0714.19.5753.4247.075320.5793.4207.0793.21.5832.4168.0832.22.5871.4129.087123.5910.4090.0910.24.5948.4052.0948The Delaware tourism board selected a simple random sample of 50 full-service restaurants located within the state. Considering all possible suchsamples with n = 50, what is the probability of selecting one whose mean is greater than 103.4? That is, p(M 103.4) =

Question

Please help with my stats hw 6. Three different distributionsThere are 559 full-service restaurants in Delaware. The mean number of seats per restaurant is 99.2. [Source: Data based on the 2002 EconomicCensus from the US Census Bureau.]Suppose that the true population mean u = 99.2 and standard deviation o = 20.5 are unknown to the Delaware tourism board. They select a simplerandom 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(Note: Although p and o are unknown tothe 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 (p = 99.2) and onerandomly selected full-service restaurant in Delaware isThe standard or typical average difference between the mean number of seats in the sample of 50 restaurants (M = 103.4) and one randomlyselected restaurant in that sample isThe standard or typical average difference between the mean number of seats in the 559 full-service restaurants in Delaware (u = 99.2) and thesample mean of any sample of size 50 isThe 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 isUse 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 thenormal table to use in determining the appropriate probability.D: Between 0and zBody BTail C 1.00 szs 1.24:1.25 szs 1.49:1.50 szs 1.74:1.75 szs 1.990.00 szs 0.24:0.50 szs 0.74:0.75 szs0.99:0.25 szs0.49:2.75 szs 2.99:3.00 szs 3.24:3.30 szS 4.00:2.00 szs 2.24:2.25 szs 2.49:2.50 szs 2.74:B: Proportion in BodyC: Proportion in TailD: Proportion Between Mean and z.5000.0000.00.5000.01.5040.4960.0040.02.5080.4920.0080.03.5120.4880.0120.04.5160.4840.0160.05.5199.4801.0199.06.5239.4761.0239.07.5279.4721.0279.08.5319.4681.0319.09.5359.4641.0359.10.5398.4602.0398.11.5438.4562.0438.12.5478.4522.0478.13.5517.4483.0517.14.5557.4443.0557.15.5596.4404.0596.16.5636.4364.0636.17.56754325.0675.18.5714.4286.0714.19.5753.4247.075320.5793.4207.0793.21.5832.4168.0832.22.5871.4129.087123.5910.4090.0910.24.5948.4052.0948The Delaware tourism board selected a simple random sample of 50 full-service restaurants located within the state. Considering all possible suchsamples 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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Given information:
The population mean and population standard deviation are 99.2 and 20.5,…

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) =