Standard distributions

Question

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The cheerleaders for a local professional basketball team must be between 58 and 80 inches. If a woman is randomly selected, what is the probability that her height is between 58 and 80 inches?

Answer 1

63.6-50=13.6=13.6/5 standard deviations from the mean or 2.52 from the mean

80-63.6=16.4=16.4/5 standard deviations from the mean or 3.28 from the mean

Using the 68-95-99.7 rule, you know over 95% of the women fall in the 50 to 80 range.

95% of the women fall within 2 standard deviations from the mean 50 and 80 > 2 standard deviations

Less than 5% will be less than 50 or greater than 80. Less than 2.5% will be less than 50, and 0.003% will be greater than 80. If rounding to 2 decimal places, a common practice, that really means no women will be taller than 80 inches, Somewhere between 2.5 and 0 is the answer. I'd guess about 1 %, about 99% of rhe women will qualify as cheerleaders.

To get a closer approximation, use Z-tables or a scientific calculator. You just need to check the lower tail of the distribution and assume no woman is too tall.All 50% of those above the mean qualify. About 49% below qualify for a grand total of about 99%. A randomly selected woman will qualify with 99% probability. Seems to be 99.4% Or you can just google the z-score and websites will calculate it for you.