Answer:
The hip breadth for men that separates the smallest 99% from the largest 1% is 17.16 inches.
Step-by-step explanation:
We are given that the Men have hip breadths that are normally distributed with a mean of 14.6 in. and a standard deviation of 1.1 in.
We have to find the hip breadth for men that separates the smallest 99% from the largest 1%.
<u><em /></u>
<u><em>Let X = length of hip breadths</em></u>
SO, X ~ Normal()
The z-score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = mean hip breadth = 14.6 inches
= standard deviation = 1.1 inches
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
<u>Now, we have to find the hip breadth for men that separates the smallest 99% from the largest 1%, which means;</u>
P(X > x) = 0.01 {where x is the required hip breadth}
P( > ) = 0.01
P(Z > ) = 0.01
<em>So, the critical value of x in the z table which represents the largest 1% of the area is given as 2.3263, that is;</em>
<em> </em>
<em> </em> = 14.6 + 2.55893 = <u>17.16 inches</u>
Hence, the hip breadth for men that separates the smallest 99% from the largest 1% is 17.16 inches.