Question 1197473: In a survey of women in a certain country (ages 2029), the mean height was 64.2 inches with a standard deviation of 2.85 inches. Answer the following questions about the specified normal distribution.
(a) What height represents the 95th percentile?
(b) What height represents the first quartile?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
I'll assume your teacher allows spreadsheet software to be used. If not, then please let me know so I can update this answer.
Open your favorite spreadsheet program. Type the command =NORMINV(0.95,64.2,2.85)
Don't forget about the equal sign up front.
The template of that function is
NORMINV(area, mu, sigma)
where,
area = value between 0 and 1
mu = mean
sigma = standard deviation
The "area" refers to the area under the curve to the left of the critical value.
The result of that command is roughly 68.8878328368117
Let's round that to four decimal places to get 68.8878
If your teacher requires some other level of precision, then be sure to use it.
About 95% of the population of women of that country, ages 20-29, are shorter than a height of 68.8878 inches.
68 inches = 60 in + 8 in = 5 ft + 8 in
Answer: 68.8878 (approximate)
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Part (b)
This time we type in =NORMINV(0.25,64.2,2.85) to find the cutoff between the lower 25% and the upper 75%
The result is roughly 62.2777042119412 which rounds to 62.27770 when rounding to five decimal places.
About 25% of the women are below a height of 62.27770 inches
62 inches = 60 in + 2 in = 5 ft + 2 in
Answer: 62.27770 (approximate)
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