Door Peephole Standing eye heights of women are normally distributed with a mean
ID: 3127381 • Letter: D
Question
Door Peephole Standing eye heights of women are normally distributed with a mean of 1516 mm and a standard deviation of 63 mm (based on anthropometric survey data from Gordon, Churchill, et al.).a. A door peephole is placed at a height that is uncomfortable for women with standing eye heights greater than 1605 mm. What percentage of women will find that height uncomfortable? b. In selecting the height of a door peephole, the architect wants its height to be suitable for the highest 99% of standing eye heights of women. What standing eye height of women separates the highest 99% of standing eye heights from the lowest 1%?
Explanation / Answer
Let X be the random variable that womens height.
X ~ Normal (mean = 1516 mm, sd = 63 mm)
A door peephole is placed at a height that is uncomfortable for women with standing eye heights greater than 1605 mm.
What percentage of women will find that height uncomfortable?
That is here we have to find P(X > 1605).
Convert x = 1605 into z-score.
z = (x - mean) / sd
z = (1605 - 1516) / 63 = 1.4127
That is now we have to find P(Z > 1.4127)
This probability we can find by using EXCEL.
syntax is,
=1 - NORMSDIST(z) (EXCEL always gives left tail probability)
where z is test statistic value = 1.4127
P(Z > 1.4127) = 0.0789
percentage = 0.0789*100 = 7.89%
The percentage of women will find that height uncomfortable is 7.89%.
In selecting the height of a door peephole, the architect wants its height to be suitable for the highest 99% of standing eye heights of women.
That is here we have given probability 0.99 and we have to find x-value.
We can find this by using EXCEL syntax is,
=NORMINV(probability, mean, standard_dev)
where probability = 0.99
mean = 1516
standard_dev = 63
x = 1662.56
What standing eye height of women separates the highest 99% of standing eye heights from the lowest 1%?
In this case also we have to find x when probability is 1% or 0.01
x = 1369.44
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