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.
a. find the probability that an individual distance is greater than 212.50 cm.
b. find the probability that the mean for 25 randomly selected distances is greater than 198.70 cm.
c. why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

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.