Question

The overhead reach distances of adult females are normally distributed with a mean of 200 cm and a standard deviation of 8.9 cm.

Answer 1

a.

$z = \dfrac{ 212.50 - 200}{8.9} = 1.4$

Looking up 1.4 in the standard normal table,

$P(z>1.4) = 1 - .9192 = 0.0808$

Answer: 8.1%

b.

The standard deviation of the average is the standard deviation of an individual sample divided by the square root of n.

$z = \dfrac{198.70 - 200}{ 8.9/\sqrt{25} } = - .7303$

$P(z > -.7303) = 1 - P( z> .7303) = P(z < .7303) = .7674$

Answer: 76.7%

c.

We know the data is normal so so is its sum and average. The standard deviation of the individual observations is given so the standard deviation of an average of n is easily calculated. We use t tests when we're estimating the standard deviation from the data.