Help please!! 3. 6.32/11.11 points | Previous Answers IllowskyintroStat1 7.HW.06

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Help please!!

3. 6.32/11.11 points | Previous Answers IllowskyintroStat1 7.HW.066. In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 58 acres. Suppose we randomly survey 48 farmers from 1940 Part (a) In words, define the random variable X the number of acres of a U.S. farm the number of animals on a U.S. farm the number of U.S. farms the number of square feet of a U.S. farm Correcti The random variable of interest is the size of a U.S. farm, in acres. Part (b) In words, define the random variable X o the average number of farms in each state in the United States the average number of animals on a U.S. farm the average number of square feet of a farm in the sample of 48 farmers the average number of acres of a farm in the sample of 48 farmers o Correct! The random variable of interest is the size of a farm, in acres. u Part (c) Give the distribution of X. (Round your standard deviation to two decimal places.) 174 Part (d) The middle 50% of the distribution for the bounds of which form the distance represented by the IQR, lies between what two values? (Round your answers to two decimal places.) acres (smaller value) acres (larger value)

Explanation / Answer

c) std deviation of sample mean =population std deviation/(n)1/2 =58/(48)1/2 =8.37

d) for middle 50% values ; z score =0.6745

therefore smaller value =sample mean -z*std deviation =174-0.6745*8.37=168.35 (approx value is 168.39

larger value =sample mean +z*std deviation =174+0.6745*8.37=179.65 (approx value is 179.61

please revert

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