Question

The height measurements of 600 adult males are arranged in ascending order and it is observed that 180th and 450th entries are 46.3 and 67.8 inch respectively. If the measurements are normally distributed. i. Find the mean and the standard deviation of the distribution. ii. Find the number of males whose height is between 40 and 70 inches.

I will be very grateful to you if you explain the solution well

Answer 1

180th entry = 180/600 x 100 = 30th percentile

450th entry = 450/600 x 100 = 75th percentile

i. Normal distribution: P(X < A) = P(Z < (A - )/)

P(X < 46.3) = 0.3

P(Z < (46.3 - )/) = 0.3

Take the value of Z corresponding to 0.3 from standard normal distribution table

(46.3 - )/ = -0.52

= 46.3 + 0.52 -------------------(1)

P(X < 67.8) = 0.75

P(Z < (67.8 - )/) = 0.75

Take the value of Z corresponding to 0.75 from standard normal distribution table

(67.8 - )/ = 0.67

= 67.8 - 0.67 -------------------(2)

From (1) and (2)

46.3 + 0.52 = 67.8 - 0.67

1.19 = 21.5

= 18.07 in

= 46.3 + 0.52x18.07 = 55.70 in

Mean, = 55.70 in and standard deviation, = 18.07 in

ii. P(40 < X < 70) = P(X < 70) - P(X < 40)

= P(Z < (70 - 55.70)/18.07) - P(Z < (40 - 55.70)/18.07)

= P(Z < 0.79) - P(Z < -0.87)

= 0.7852 - 0.1922

= 0.5930

Number of males whose height is between 40 and 70 inches = 0.5930 x 600

= 355.8

= 356

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