Question

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 69 inches and standard...

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 69 inches and standard deviation 1 inch.

If a random sample of thirty 18-year-old men is selected, what is the probability that the mean height x is between 68 and 70 inches? (Round your answer to four decimal places.)

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Answer #1

Solution :

Given that,

mean = \mu = 69

standard deviation = \sigma = 1

n = 30

\mu\bar x = \mu = 69

\sigma\bar x = \sigma / \sqrt n = 1 / \sqrt 30 = 0.1826

P( 68< \bar x < 70) = P((68 - 69) / 0.1826 <(\bar x - \mu \bar x) / \sigma \bar x < (70 - 69) / 0.1826))

= P(-5.48 < Z < 5.48 )

= P(Z < 5.48) - P(Z < 5.48) Using standard normal table,  

= 1 - 0

= 1

Probability = 1.0000

The probability that the mean height x is between 68 and 70 inches is 1.0000.

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