For women age 18-24 in a national study, the average height was about 64.3 inches with an SD of about 2.6 inches. The heights were approximately normally distributed. If a woman is selected at random from the study, what is the chance that she is between 60 and 66 inches

69.27% probability that she is between 60 and 66 inches

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

If a woman is selected at random from the study, what is the chance that she is between 60 and 66 inches

This is the pvalue of Z when X = 66 subtracted by the pvalue of Z when X = 60. So

X = 66

has a pvalue of 0.7422

X = 60

has a pvalue of 0.0495.

0.7422 - 0.0495 = 0.6927

69.27% probability that she is between 60 and 66 inches

P(60 ≤ x ≤ 66) = 0.69

Step-by-step explanation:

Since the heights were approximately normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = heights of the women

µ = mean height

σ = standard deviation

From the information given,

µ = 64.3 inches

σ = 2.6 inches

The the chance that she is between 60 and 66 inches is expressed as

P(60 ≤ x ≤ 66)

For x = 60,

z = (60 - 64.3)/2.6 = - 1.65

Looking at the normal distribution table, the probability corresponding to the z score is 0.05

For x = 66,

z = (66 - 64.3)/2.6 = 0.65

Looking at the normal distribution table, the probability corresponding to the z score is 0.74

Therefore,

P(60 ≤ x ≤ 66) = 0.74 - 0.05 = 0.69