A survey found that women\'s heights are normally distributed with mean 62.6 in.
ID: 3151359 • Letter: A
Question
A survey found that women's heights are normally distributed with mean 62.6 in. and standard deviation 2.7 in. The survey also found that men's heights are normally distributed with a mean 68.1 in. and standard deviation 2.7. Complete parts a through c below. a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 2 in. Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is %. (Round to two decimal places as needed.) b. Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is %. (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.) Show work
Explanation / Answer
a)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 57
x2 = upper bound = 74
u = mean = 62.6
s = standard deviation = 2.7
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2.074074074
z2 = upper z score = (x2 - u) / s = 4.222222222
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.019036216
P(z < z2) = 0.999987905
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.980951689 and 98.0951689% [ANSWER]
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b)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 57
x2 = upper bound = 74
u = mean = 68.1
s = standard deviation = 2.7
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -4.111111111
z2 = upper z score = (x2 - u) / s = 2.185185185
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 1.9688E-05
P(z < z2) = 0.985562364
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.985542676 = 98.5542676% [ANSWER]
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c)
FOR THE LOWER BOUND:
First, we get the z score from the given left tailed area. As
Left tailed area = 0.05
Then, using table or technology,
z = -1.644853627
As x = u + z * s,
where
u = mean = 62.6
z = the critical z score = -1.644853627
s = standard deviation = 2.7
Then
x = critical value = 58.15889521
FOR THE UPPER BOUND:
First, we get the z score from the given left tailed area. As
Left tailed area = 0.95
Then, using table or technology,
z = 1.644853627
As x = u + z * s,
where
u = mean = 68.1
z = the critical z score = 1.644853627
s = standard deviation = 2.7
Then
x = critical value = 72.54110479
Hence,
it is between 58.15889521 and 72.54110479. [ANSWER]
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