Suppose x has a distribution with mu = 16 and sigma = 9. (a) If a random sample

Question

Suppose x has a distribution with mu = 16 and sigma = 9. (a) If a random sample of size n = 47 is drawn, find mu _x^bar, sigma _x^bar and P(16 lessthanorequalto x^bar lessthanorequalto 18). (Round sigma _x^bar to two decimal places and the probability to four decimal places.) mu _x^bar = ______ sigma _ x^bar = _____ P(16 lessthanorequalto x^bar lessthanorequalto 18) = ________ (b) If a random sample of size n = 63 is drawn, find mu _x^bar, sigma _x^bar and P(16 lessthaorequato x^bar lessthaorequato 18). (Round sigma_ x^bar to two decimal places and the probability to four decimal places.) mu _x^bar = ______ sigma _x^bar = _____ P(16 lessthaorequato x^bar lessthaorequato 18) = _______ Why should you expect the probability of part (b) to be higher than that of part (a)?

Explanation / Answer

a)
x = 16
x = 9/sqrt(47) = 1.312

Mean = 16
Standard deviation = 9
Standard error / n = 9 / 47 = 1.3127

standardize xbar to z = (xbar - ) / ( / n )
P( 16 < xbar < 18) = P[( 16 - 16) / 1.3127 < z < ( 18 - 16) / 1.3127]
P( 0 < z < 1.5235) = 0.0643
(from normal probability table)

b)
Mean = 16
Standard deviation = 9
Standard error / n = 9/ 63 = 1.1338
standardize xbar to z = (xbar - ) / ( / n )
P( 16 < xbar <18) = P[( 16- 16) / 1.1338 < z < ( 18 - 16) / 1.1338]
P( 0 < z < 1.1338) = 0.1292
(from normal probability table)

c)
The standard deviation (of the mean) is smaller in (b) ; dividing by a smaller number makes the area larger.
Therefore, the distribution about x is wider.

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