Suppose the heights of 18-year-old men approximately normally distributed, with

Question

Suppose the heights of 18-year-old men approximately normally distributed, with mean 71 ides and standard dniation 3 inches. (e) What is the probability that an 18-year-old man selected at randorn is between 70 and 72 inches tall? (Round your answer to four decimal places) ra random sample oft very 18-year-old men is selected what the pot at ity that themeen he ???.bet een ond anos, e und ye meter pa (c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this? O The probability in part (b) is much higher because the standard deviation is larger for the x dstributiorn O The probability in part (b) is much higher because the mean is smaller for the x dstribution The probablity in part (b) is much higher because the standard deviation imeter for thai datribution O The probability in part (b) is much higher because the mean is larger for the i distritution O The probability in part (b) is much lower because the standard de

Explanation / Answer

a) mean= 71 , sd = 3
a)
P(70 <X< 72)
Z = (X - 71)/3
=P ( ?0.33<Z<0.33 )=0.2586

b)
Xbar follow N( mean, sd^2/n)
= N(71 , 3^2/20)

Z = (Xbar - 71)/0.67082039325

P(70 <Xbar< 72)
= P ( ?1.49<Z<1.49 )=0.8638

c)

option C) is correct as sd is smaller here

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