-/9 points BBBasicStat7 7.5.011 My Notes Suppose x has a distribution with 12 an
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-/9 points BBBasicStat7 7.5.011 My Notes Suppose x has a distribution with 12 and = 7. (a) If a random sample of size n = 43 is drawn, find , and P(12 x 14). (Round CF to two decimal places and the probability to four decimal places.) (b) If a random sample of size n = 68 is drawn, find , and P(12 x š 14). (Round ox to two decimal places and the probability to four decimal places.) (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 The standard deviation of part (b) is Select- part (a) because of the sample size. Therefore, the distribution about is-select Selec -- Need Help? Read It Talk to a Tutor -4 points BBBasicStat7 7.5.013 My NotesExplanation / Answer
a. Pop mean and sigma = 12, 7 n=43,
Sample mean and Sample dev = 12, and 7/sqrt(n) = 7/sqrt(43) = 1.0675 respectively
So, in first and second space put: 12, 1.0675
P(12<=X<=14) = P(12-12 / 1.0675 <=Z<= 14-12/1.0675)
= P(0<=Z<=1.873) = .9693 - .5 = .4693
In 3rd space put = .4693
b. Similarly,
In 1st and 2nd space put: 12, 7/sqrt(68) = .8489
P(12<=Xbar<=14) = P(12-12/.8489<=Z<= 14-12/.8489)
=P(0<=Z<=2.3561) = .9909-.5 = .4909
In 3rd space out : .4909
c. When you sample size is greater ( it turns from 43 to 68 in our case), the sampling distribution transforms to normal distribution. In normal distribution, more area is centered around mean than the sampling distribution. Hence, the area we get for b) is more than area in part a). In other words standard deviation of b) is less than a) hence, covering more area for the same Xbar range.
The standard deviation of part (b) is less than part (a) because of the larger sample size. Therefore, the distribution about MeanBar is ( I don't know what options are here - guesses - more/thicker/denser). But hope you get the context of what i am trying to explain.
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