Suppose x has a distribution with ? = 24 and ? = 17. (a) If a random sample of s

Question

Suppose x has a distribution with ? = 24 and ? = 17.

(a) If a random sample of size n = 32 is drawn, find ?x, ?x and P(24 ? x ? 26). (Round ?x to two decimal places and the probability to four decimal places.)

(b) If a random sample of size n = 69 is drawn, find ?x, ?x and P(24 ? x ? 26). (Round ?x to two decimal places and the probability to four decimal places.)

P(24 ? x ? 26) =

(c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
The standard deviation of part (b) is ________ (a) because of the ________ sample size. Therefore, the distribution about ?x is ________

possible answers: the same, larger, smaller. wider, the same, narrower

Explanation / Answer

Ans:

a)mean=24

standard dveiation=17/sqrt(32)=3.01

z(24)=(24-24)/3.01=0

z(26)=(26-24)/3.01=0.666

P(0<=z<=0.666)=P(z<=0.666)-P(z<=0)=0.7471-0.5=0.2471

*(if rounded off value of std. dev. has to be used ,the prob. will be 0.2468)

b)mean=24

standard deviation=17/sqrt(69)=2.05

z(24)=0

z(26)=(26-24)/2.05=0.977

P(0<=z<=0.977)=P(z<=0.977)-P(z<=0)=0.8358-0.5=0.3358

*(if rounded off value of std. dev. has to be used ,the prob. will be 0.3354)

c)standard deviation in part. b will be smaller(2.05) as compared to part a(3.01),as sample size is larger in part b.

Therefore, the distribution about ?x is narrower.

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