The distribution of height of adult american women is approximately normal with a mean of 65.5 inches and a standard deviation o
f 2.5 inches what percentage of women are taller than 70.5 inches?
1 answer:
Answer:
2.275%
Step-by-step explanation:
We start by calculating the z-score
Mathematically;
z-score = (x-mean)SD
here, x = 70.5
mean = 65.5
SD = 2.5
Substituting these values;
z = (70.5-65.5)/2.5
= 5/2.5 = 2
So we want to calculate the probability that;
P( z> 2)
We check the standard normal distribution table for this
That will be
0.02275
In percentage, this is 2.275 %
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