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A survey found that women's heights are normally distributed with mean 62.8 inches and standard deviation of 3.7 inches. This survey also found that men's heights are normally distributed with a mean 68.1 and a standard deviation 3.5.
Most of the live characters employed at an amusement park have height requirements of a minimum 57in a d a maximum 64in.
A.Find the percentage of men meeting the height requirement
B. If the height requirements are changed to exclude only the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

Respuesta :

Using the normal distribution, it is found that:

a) The percentage of men meeting the height requirement is 12.02%.

b) The new height requirements are between 62.3 and 68.1 in.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • Men have a mean height of 68.1 inches, thus [tex]\mu = 68.1[/tex].
  • Standard deviation of 3.5 in, thus [tex]\sigma = 3.5[/tex].

Item a:

The proportion between 57 in and 64 in is the p-value of Z when X = 64 subtracted by the p-value of Z when X = 57, thus:

X = 64:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{64 - 68.1}{3.5}[/tex]

[tex]Z = -1.17[/tex]

[tex]Z = -1.17[/tex] has a p-value of 0.1210.

X = 57:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{57 - 68.1}{3.5}[/tex]

[tex]Z = -3.17[/tex]

[tex]Z = -3.17[/tex] has a p-value of 0.0008.

0.1210 - 0.0008 = 0.1202.

0.1202 x 100% = 12.02%

The percentage of men meeting the height requirement is 12.02%.

Item b:

  • The new heights requirements would be between the 5th and the 50th percentile.
  • The 5th percentile is X when Z has a p-value of 0.05, so X when Z = -1.645.
  • The 50th percentile is X when Z has a p-value of 0.5, so X when Z = 0.

Then:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.645 = \frac{X - 68.1}{3.5}[/tex]

[tex]X - 68.1 = -1.645(3.5)[/tex]

[tex]X = 62.3[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0 = \frac{X - 68.1}{3.5}[/tex]

[tex]X - 68.1 = 0(3.5)[/tex]

[tex]X = 68.1[/tex]

The new height requirements are between 62.3 and 68.1 in.

A similar problem is given at https://brainly.com/question/25154104

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