A survey found that women\'s heights are normally distributed with mean 62.6 in

Question

A survey found that women's heights are normally distributed with mean 62.6 in and the standard deviation 2.3. The survey also found that men's heights are normaly distributed with a mean 69.3 and standard deviation 2.9 A) most characters at an amusement park have height requirements with a mininmum of 4 ft 9 in. and a maximum of 6 ft 3 in. Find the percentage of women meeting the requirement. The percentage of women who meet the height requirement is % B) The percentage of men who meet the requirement is % C) If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% or women what are the new height requirements? The new height requirements are at least in. and at most in.

Explanation / Answer

A) From information given, Xbar=62.6 in, s=2.3, Find Z scores corresponding to Xi=57 in (4ft9in=4*12+9=57in) and 75 in.

Z1=(Xi-Xbar)/s=(57-62.6)/2.3=-2.43 and Z2=(75-62.6)/2.3=5.39

P(57<X<75)=0.4925+0.4999=0.9924~99.24% (ANS)

B) Z scores corresponding to men's height

Z1=(57-69.3)/2.9=-4.24 and Z2=(75-69.3)/2.9=1.97

P(57<X<75)=0.4999+0.4756=0.9755

C) Find Xi corresponding to -1.64 and 1.64

-1.64=(Xi-62.6)/2.3

Xi=62.6-3.772=58.828

1.64=(Xi-69.3)/2.9

Xi=74.056

Height is atleast 58.828 and atmost 74.056.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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