We are given that the average height of 20-year-old American women is normally distributed with a mean of 64 inches and standard deviation of 4 inches.

Also, a random sample of 4 women from this population is taken and their average height is computed.

Let X bar = Average height

The z score probability distribution for average height is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 64 inches

$\sigma$ = standard deviation = 4 inches

n = sample of women = 4

So, Probability that average height would be shorter than 63 inches is given by = P(X bar < 63 inches)

P(X bar < 63) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{63-64}{\frac{4}{\sqrt{4} } }$ ) = P(Z < -0.5) = 1 - P(Z <= 0.5)

= 1 - 0.69146 = 0.30854

Hence, it is 30.85% likely that average height would be shorter than 63 inches.

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4 years ago

Simplify (45-28) - |15-21| A) 11 B) 17 C) 23 D) 29​

Simplify (45-28) - |15-21| A) 11 B) 17 C) 23 D) 29