women's heights are normally distributed with a mean 63.1 in and standard deviation of 2.5. A social organization for tall people has a requirement that women must be at least 69in tall. What percentage of woman meet that requirement?

The percentage of woman that are taller than 69in is __%

z = (69-63.1)/ 2.5

z = 5.9/2.5
z = 2.36

z = (69-63.1)/ 2.5

z = 5.9/2.5
z = 2.36

Once you have z-scores, check a z-table for the probability scores. Convert to a percentage.

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Assume that​ women's heights are normally distributed with a mean given by mu equals 62.3in​, and a standard deviation given by sigma equals 2.7

​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 6363 in.
​(b) If 43 women are randomly​ selected, find the probability that they have a mean height less than 63 in.

To find the percentage of women who meet the height requirement of being at least 69 inches tall, we need to calculate the probability of a woman's height being greater than or equal to 69 inches.

Since the heights of women are normally distributed with a mean of 63.1 inches and a standard deviation of 2.5 inches, we can use the z-score formula to standardize the height value.

The z-score is calculated by subtracting the mean from the given value (69 inches in this case) and dividing it by the standard deviation.

z = (x - μ) / σ

So, z = (69 - 63.1) / 2.5 = 2.36

Now, we need to find the percentage of women who have a z-score greater than or equal to 2.36.

Using a standard normal distribution table or a calculator, we can find the corresponding area (probability) for this z-score. This area represents the percentage of women who have a height greater than or equal to 69 inches.

Looking up the z-score of 2.36 in the standard normal distribution table, we find the area to be approximately 0.9900.

To convert this to a percentage, we multiply by 100.

0.9900 * 100 = 99.00%

Therefore, approximately 99% of women meet the height requirement of being at least 69 inches tall.