In a study of female students who suffer from bulimia, each student completed a

Question

In a study of female students who suffer from bulimia, each student completed a questionnaire from which a "fear of negative evaluation" (FNE) score was produced. (The higher the score, the greater is the fear of negative evaluation.) Suppose the FNE scores of bulimic students have a distribution with mean mu = 24 and standard deviation sigma = 5. Now, consider a random sample of 35 female students with bulimia. Complete parts a through c below. a. What is the probability that the sample mean FNE score is greater than 22.5? (Round to four decimal places as needed.) b. What is the probability that the sample mean FNE score is between 24 and 25.6? (Round to four decimal places as needed.) c. What is the probability that the sample mean FNE score is less than 25.6? (Round to four decimal places as needed.)

Explanation / Answer

mean = 24 , s = 5 , n = 35
a) P(X > 22.5)
z = ( x - mean) /( s / sqrt(n))
= ( 22.5 - 24 ) /( 5 / sqrt(35))
= -1.7748
we need to find P(z > -1.7748)
P(X > 22.5) = P(z > -1.7748) = 0.9621

b) P(24 < X < 25.6)
P(X < 24)
z = ( x - mean) /( s / sqrt(n))
= ( 24 - 24 ) /( 5 / sqrt(35))
= 0

P(X < 25.6)
z = ( x - mean) /( s / sqrt(n))
= ( 25.6 - 24 ) /( 5 / sqrt(35))
= 1.8931
P(24 < X < 25.6) = P(0 < z < 1.8931) = 0.4709

c)
P(X < 25.6)
z = ( x - mean) /( s / sqrt(n))
= ( 25.6 - 24 ) /( 5 / sqrt(35))
= 1.8931
we need to find P(z < 1.8931)
P(X < 25.6) = P(z < 1.8931) = 0.9709

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