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 65.5 and 68.0 inches. If a woman is randomly selected, what is the probability that her height is between 65.5 and 68.0 inches

The probability that the height of a selected woman's height is between 65.5 and 68.0 is 0.21

Step-by-step explanation:

Assume that the heights of women are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = heights women.

µ = mean height

σ = standard deviation

From the information given,

µ = 63.6 inches

σ = 2.5 inches

The probability that the height of a selected woman's height is between 65.5 and 68.0 inches is expressed as

P(65.5 ≤ x ≤ 68)

For x = 65.5,

z = (65.5 - 63.6)/2.5 = 0.76

Looking at the normal distribution table, the probability corresponding to the z score is 0.7764

For x = 68,

z = (68 - 63.6)/2 = 2.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.986

Therefore,

P(65.5 ≤ x ≤ 68) = 0.986 - 0.7764 = 0.21

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

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step-by-step explanation:

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