Question

For #4-5, use the following information: Overhead reach distances X are used in planning assembly workstations. The overhead reach distance of adult females is assumed to be X ~ N (202.5 cm, 8.0 cm) where cm refers to centimeters.

4. If 1 adult female is randomly selected, find the probability that her overhead reach is between 194.5 cm and 210.5 cm. Use the graph to sketch the probability as the area under the pdf given.

P(X < x) n X X

5. If 64 adult females are randomly selected, find the probability that they have a mean overhead reach between 200.5 cm and 204.5 cm. Use the graph to sketch the probability as the area under the pdf given.

P(X < x) x X

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Answer #1

4.

The following information has been provided:

μ=202.5, σ=8

We need to compute Pr(194.5≤X≤210.5).

The corresponding z-values needed to be computed are:

X1 - 194.5 – 202.5 21 -1 0 8

Z_2 = \frac{X_2 - \mu}{\sigma} = \frac{ 210.5 - 202.5}{ 8}= 1

Therefore, we get:

\Pr(194.5 \leq X \leq 210.5) = \Pr\left(\frac{ 194.5 - 202.5}{ 8} \le Z \le \frac{ 210.5 - 202.5}{ 8}\right)

= \Pr(-1 \le Z \le 1)

= \Pr(Z \le 1) - \Pr(Z \le -1) = 0.8413 - 0.1587 = 0.6827

Normal Distribution: Pr(194.5<X<210.5)=0.6827 0.050 0.045 0.040 0.035 0.030 0.025 0.020 0.015 0.010 0.005 0.000 170 175 180 1

5.

The following information about the mean and standard deviation has been provided:

μ=202.5, σ=8, n=64

We need to compute Pr(200.5≤Xˉ≤204.5).

The corresponding z-values needed to be computed are:

Z_1 = \frac{\bar X_1 - \mu}{\sigma/\sqrt{n}}= \frac{ 200.5-202.5}{ 8/\sqrt{ 64}} = -2

Z_2 = \frac{X_2-\mu}{\sigma/\sqrt{n}} = \frac{ 204.5-202.5}{ 8/\sqrt{ 64}}= 2

Therefore, the following is obtained:

\Pr(200.5 \leq \bar X \leq 204.5) = \Pr\left(\frac{ 200.5-202.5}{ 8/\sqrt{ 64}} \le Z \le \frac{ 204.5-202.5}{ 8/\sqrt{ 64}}\right)

= \Pr(-2 \le Z \le 2)

= \Pr(Z \le 2)-\Pr(Z \le -2)

= 0.9772-0.0228 = 0.9545

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