2018 Spring MA Homework: Section 8.1 Score: 3.33 of 10 pts Kaysha Adams&1 2/18 1

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2018 Spring MA Homework: Section 8.1 Score: 3.33 of 10 pts Kaysha Adams&1 2/18 1022 PM Homework Hw Score: 26%, 19.5 of 75 pts a Assigned Meda Question Help A simple random sample of size n-74 is obtained from a population with dstribution of x t 66 and a 3. Does the o be approximabely normalily distributed? Why? What is the sampling distributlion af need to be normally distbuted for the sampling of x to be approximalely normally distributod? Why? A Yes becauae the Central Limit Theorem staes hat only for undertying pepuletions that are nomal is the shape of the sampling distribution of i nomal, O8. Yes because the Central Limit Thacrem staos that the sampling variability of nonnomal pepuations w egardiess of the sample size, n wil increase as the sample size increasos No because the Central Linit Theorem states that only i the shape of the undertying popuiation is normal or unfom beoome approximately nomal as the sample sice, m, increases does the sampling distribution of x D. No because the Central Limit Theorem states that egandless of the shape of the underying population, the sampling dstribtion of x becomes approximaely normal as the sample sine, n inreases choice below places as meeded) boxes withn your choice. Type integers or decimals rounded to three deoimal OA. The sampling distribution of i folows Stuadent's t dstibution with O B. The smpling distribution of i is uniform with ande O G. Thesapling distribution of i is shewed iwt with y and and Click to select and enter your answeris) and then cick Check Answer All parts showing Clear All 2 5

Explanation / Answer

We have sample size n = 74

The population mean and SD is ?= 66 and ? = 3

As per the CLT, irrespective of the actual distribution, if the sample size is above 30, it behaves as normal distribution

So, answer for the first question will be - (D)

So for x-bar, the distribution is normal

Mean = 66

SD = ?/sqrt(n) = 3/sqrt(74) = 0.348

Hope this helps and Please don't forget to rate the answer :)

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