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Margarita [4]
3 years ago
5

A survey found that​ women's heights are normally distributed with mean 63.3 in. and standard deviation 2.7 in. The survey also

found that​ men's heights are normally distributed with a mean 67.3 in. 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 6ft 4in find the percentage of women meeting the height requirement
the percentage of woment who meet the height requirement?
(round to two decimal places as needed)
b) find the percentage of men meeting the height requirement
the percentage of men meeting the height requirement
(round to two decimal places as needed )
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 ___ in. and at most ___ in.
(round to one decimal place as needed)
Mathematics
1 answer:
mylen [45]3 years ago
5 0

Answer:

A) There is a 100% chance of meeting the height requirement for women.

B) There is a 99.89% chance of meeting the height requirement for men.

C) New height requirements are:

71.9 inches for men and 58.9 inches minimum for women.

At least 58.9 inches and at most 71.9 inches.

Step-by-step explanation:

This question involves finding z values which tell us the percentage from 0 to x.

z = (x-μ)/σ

A) In this question, the minimum sample point is 4 ft 9 inches or converting to inches, we have;(4*12) + 9 =57 inches.

The maximum sample point is 6 ft 4 inches which in inches gives; (6*12) + 4 = 76 inches

For the minimum point;

z = (57 - 63.3)/2.7 = -2.33

From the z-distribution table, this equates to 0.0099.

The maximum;

z = (76 - 63.3)/2.7 = 4.70

From the z-table, it gives 0.999.. Basically 100%.

So there is a 100% chance of meeting the height requirement for women.

B) For men;

the minimum z = (57 - 67.3)/2.8 = -3.68 which gives 0.00012 from the z-distribution table.

The maximum;

z = (76 - 67.3)/2.8 = 3.07 which gives 0.9989 from the z-distribution table. Basically, 99.89%.

So there is a 99.89% chance of meeting the height requirement for men.

C) To exclude the tallest 5% of men, we need to find the z-value for 0.95 and then solve for x.

Thus from the z-table, z = 1.65. So;

1.645 = (x - 67.3)/2.8

4.606 = x - 67.3

x = 67.3 + 4.606

x ≈ 71.9 inches for men.

So, Current minimum is okay.

To exclude the shortest 5% of women, we need to find the z for 0.05.

From the z-distribution table, it has a value of -1.645. So;

-1.645 = (x - 63.3)/2.7

-4.4415 = x - 63.3

x = 63.3 - 4.4415

x ≈ 58.9 inches minimum.

So, Current maximum is okay.

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The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118   .

In the question ,

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