3) Disney World requires that people employed as mickey mouse characters must have a height between 56in and 62 in. Suppose mens heights are normally dis- tributed with a mean of 68.6in and a standard deviation of 2.8 in. Suppose women heights are normally distributed with a mean of 63.7in and a standard deviation of 2.9 in.a) Find the percentage of men meeting the height requirement.b) If the height requirement is changed to exclude the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

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Asked Dec 5, 2019
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3) Disney World requires that people employed as mickey mouse characters must have a height between 56in and 62 in. Suppose mens heights are normally dis- tributed with a mean of 68.6in and a standard deviation of 2.8 in. Suppose women heights are normally distributed with a mean of 63.7in and a standard deviation of 2.9 in.

a) Find the percentage of men meeting the height requirement.

b) If the height requirement is changed to exclude the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

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Expert Answer

Step 1

a)

The percentage of men meeting the height requirement is obtained below:

From the given information, the mean value for men’s heights is 68.6 in and the standard deviation value for men’s heights is 2.8 in. the mean value for women heights is 63.7 in and the standard deviation value for women heights is 2.9 in.

The required probability is,

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a-u X-u b-u P(a< X <b)= P 56-68.6 X-68.6 62- 68.6 P(56 < X < 62) = P 2.8 2.8 2.8 -12.6 -6.6 2.8 2.8 = P(-4.50 <z<-2.36) = P(z<-2.36) – P(z<4.50)

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Step 2

From the “standard normal table”, the area to the left of Z = –2.36 is 0.0091, and the area to the left of Z = –4.50 is 0.0000.

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Р (56 < X <62) - P(г<-2.36)- Р(z < -450) = 0.0092 – 0.0000 = 0.0092

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Step 3

b)

The tallest 50% of men new height requirement is obtained below:

From the given information, the mean value for men’s heights is 68.6 in and the standard deviation value for men’s heights is 2.8 in. the tallest 50% of the data values are larger than cutoff values and thus, 100% – 50% = 50%  of the data values are smaller than the cutoff value.

The positive z value is obtained below:

Procedure for finding the z-value is listed below:

  1. From the table of standard norm...
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x = u+ zo = 68.6 +(0.00)(2.8) = 68.6

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