Question

The overhead reach distances of adult females are normally distributed with a mean of 205.5 cm and a standard deviation of 8.9 cm. a. Find the probability that an individual distance is greater than 215.50 cm. b. Find the probability that the mean for 20 randomly selected distances is greater than 204.20 cm. c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

Answer:

a) 13.14% probability that an individual distance is greater than 215.50 cm.

b) 74.22% probability that the mean for 20 randomly selected distances is greater than 204.20 cm.

c) Because the overhead reach distances of adult females are normally distributed

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 205.5, \sigma = 8.9$

a. Find the probability that an individual distance is greater than 215.50 cm.

This is the pvalue of Z when X = 215.50. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{215.5 - 205.5}{8.9}$

$Z = 1.12$

$Z = 1.12$ has a pvalue of 0.8686

1 - 0.8686 = 0.1314

13.14% probability that an individual distance is greater than 215.50 cm.

b. Find the probability that the mean for 20 randomly selected distances is greater than 204.20 cm.

Now we have $n = 20, s = \frac{8.9}{\sqrt{20}} = 1.99$

This probability is 1 subtracted by the pvalue of Z when X = 204.20. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{204.2 - 205.5}{1.99}$

$Z = -0.65$

$Z = -0.65$ has a pvalue of 0.2578

1 - 0.2578 = 0.7422

74.22% probability that the mean for 20 randomly selected distances is greater than 204.20 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Because the overhead reach distances of adult females are normally distributed

Answer 2

Answer:

(a) P(X > 215.50) = 0.1314

(b)  P($\bar X$ > 204.20) = 0.55962

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 8.9 cm.

Let X = individual reach distances of adult females

So, X ~ N($\mu=205.5,\sigma^{2} = 8.9^{2}$)

The z score probability distribution is given by;

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

where, $\mu$ = population mean

            $\sigma$ = population standard deviation

(a) Probability that an individual distance is greater than 215.50 cm is given by = P(X > 215.50)

    P(X > 215.50) = P( $\frac{X-\mu}{\sigma}$ > $\frac{215.50-205.50}{8.9}$ ) = P(Z > 1.12) = 1 - P(Z $\leq$ 1.12)

                                                                  = 1 - 0.86864 = 0.1314

(c) Now as we are given with the sample of 20 randomly selected distances . So, the z score probability distribution for sample mean is given as;

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

where, $\mu$ = population mean

           $\sigma$ = population standard deviation

           n = sample size = 20

So, Probability that the mean for 20 randomly selected distances is greater than 204.20 cm = P($\bar X$ > 204.20 cm)

    P($\bar X$ > 204.20) = P( $\frac{X-\mu}{\sigma}$ > $\frac{204.20-205.50}{8.9}$ ) = P(Z > -0.15) = P(Z < 0.15)

                                                                   = 0.55962

(c) Normal distribution can be used in part​ (b), even though the sample size does not exceed​ 30 because we are given in the question that the distances of adult females are normally distributed.