Assume heights of women in a population follow the normal curve with mean 64.3 i

Question

Assume heights of women in a population follow the normal curve with mean 64.3 inches and standard deviation 2.6 inches.

A) What proportion of women have heights between 60 and 66 inches?

B) A certain woman has a height 0.5 standard deviations above the mean. What proportion of women are taller than she is?

C) How tall is a woman whose height is on the 90th percentile?

D) A women is chosen at random from this population. What is the probability that she is more than 67 inches tall?

E) Five women are chosen at random from this population. What is the probability that exactly one of them is more than 67 inches tall?

Please answer ALL of the questions, and do not copy and paste from the other answers out there, because those other answers out there are not complete.

Explanation / Answer

a) P( 60 < X < 66) = P((60 - mean)/sd < (X - mean)/sd < (66 - mean)/sd

= P((60 - 64.3)/2.6 < Z < (66 - 64.3)/2.6)

= P(-1.65 < Z < 0.65)

= P(Z < 0.65) - P(Z < -1.65)

= 0.7422 - 0.0495 = 0.6927

b) P(X > 64.3 + 0.5) = P(X > 64.8)

    = P((X - mean)/sd > (64.8 - mean)/sd)

    = P(Z > (64.8 - 64.3)/2.6)

     = P(Z > 0.19)

      = 1 - P(Z < 0.19)

      = 1 - 0.5753 = 0.4247

c) P(X < x) = 0.9

or, P((X - mean)/sd < (x - mean)/sd) = 0.9

or, P(Z < (x - 64.3)/2.6) = 0.9

or, (x - 64.3)/2.6 = 1.28

or, x = 1.28 * 2.6 + 64.3

or, x = 67.628

d) P(X > 67) = P(Z > (67 - 64.3)/26)

                     = P(Z > 0.1)

                     = 1 - P(Z < 0.1)

                     = 1 - 0.5398 = 0.4602

e) P(X > 67) = P((X - mean)/(sd/sqrt(n)) > (67 - mean)/(sd/sqrt(n))

                     = P(Z > (67 - 64.3)/(2.6/sqrt(5))

                     = P(Z > 2.32)

                     = 1 - P(Z < 2.32)

                     = 1 - 0.9898 = 0.0102

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.