Question

The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 cm
a. Find the probability that an individual distance is greater than 214.80 cm.
b. Find the probability that the mean for 1515 randomly selected distances is greater than 204.00 cm
c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?
a. The probability is
​(Round to four decimal places as​ needed.)
b. The probability is
​(Round to four decimal places as​ needed.)
c. Choose the correct answer below.
A. The normal distribution can be used because the original population has a normal distribution.
B. The normal distribution can be used because the probability is less than 0.5
C. The normal distribution can be used because the mean is large.
D. The normal distribution can be used because the finite population correction factor is small

Answer 1

Answer:

(a) the probability that an individual distance is greater than 214.80 cm is 0.1401.

(b) The probability that the mean for 15 randomly selected distances is greater than 204.00 cm is 0.2482.

(c) The normal distribution can be used because the original population has a normal distribution.

Step-by-step explanation:

We are given that the overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.6 cm.

(a) Let X = the overhead reach distances of adult females.

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = population mean reach distance = 205.5 cm  

           $\sigma$ = standard deviation = 8.6 cm

So, X ~ Normal($\mu=205.5,\sigma^{2} =8.6^{2}$)

Now, the probability that an individual distance is greater than 214.80 cm is given by = P(X > 214.80 cm)

    P(X > 214.80 cm) = P( $\frac{X-\mu}{\sigma}$ > $\frac{214.80-205.5}{8.6}$ ) = P(Z > 1.08) = 1 - P(Z $\leq$ 1.08)

                                                                    = 1 - 0.8599 = 0.1401

The above probability is calculated by looking at the value of x = 1.08 in the z table which has an area of 0.8599.

(b) Let $\bar X$ = the sample mean selected distances.

The z-score probability distribution for the sample mean is given by;

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

where, $\mu$ = population mean reach distance = 205.5 cm  

           $\sigma$ = standard deviation = 8.6 cm

           n = sample size = 15

Now, the probability that the mean for 15 randomly selected distances is greater than 204.00 cm is given by = P($\bar X$ > 204.00 cm)

    P($\bar X$ > 204 cm) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{204-205.5}{\frac{8.6}{\sqrt{15} } }$ ) = P(Z > -0.68) = 1 - P(Z $\leq$ 0.68)

                                                                    = 1 - 0.7518 = 0.2482

The above probability is calculated by looking at the value of x = 0.68 in the z table which has an area of 0.7518.

(c) The normal distribution can be used in part​ (b), even though the sample size does not exceed​ 30 because the original population has a normal distribution and the sample of 15 randomly selected distances has been taken from the population itself.