A survey found that womens heights are normally distributed

A survey found that women\'s heights are normally distributed with mean 62.9in. and standard deviation 2.5in. The survey also found that men\'s heights are normally distributed with a mean68.6in. and standard deviation 2.8. Complete parts A through C below.

A - Most of the live characters at an amusement park have height requirements with a minimum of 4ft 9in. and a maximum of 6 ft 2in. Find the percentage of women meeting the height requirement.

The percentage of women who meet the height requirement is______?

B - Find the percentage of men meeting the height requirement.

The percentage of men who meet the height requirement is________?

C - If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements?

The new height requirements are at least­­­______inches and at most_______inches.

Solution

X =Height of women is normal N(62.9, 2.5)

Y =Men heights (68.6, 2.8)

A) P(4\'9\'<x<6\'2\")

= P(57<x<74)

Z = (x-62.9)/2.5

Hence reqd prob=

P(-4.3/2.5 < Z<11.1/2.5)

= P(-1.72<Z<4.44)

= 0.4573+0.5= 0.9573

i.e. 95.73% of women meet the height requirement.

B) For men Z = Y-68.6/2.8

P(57<x<74)

= P(-4.143<z<1.93)

= 0.5+0.4732

=0.9732

97.32% of men in amusement park meet this requirement

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C) For tallest 5% z = 1.645

Y = 68.6+2.8(1.645)

= 73.206

For smallest 5%

x = 62.9-1.645(2.5)

=58.7875

Hence 58.7875<height<73.206 inches