Question

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 42 ounces and a standard deviation of 10 ounces. Use the Empirical Rule.
a. 99.7% of the widget weights lie between _____ and _____.
b. What percentage of the widget weights lie between 26 and 66 ounces?
c. What percentage of the widget weights lie above 34?

Answer 1

Answer:

The answer is below

Step-by-step explanation:

The empirical rule states for a normal distribution, 68% of the data falls within one standard deviation, 95% falls within two standard deviations and 99.7% falls within three standard deviations.

z score is given by:

$z=\frac{x-\mu}{\sigma}$

Given that mean (μ) = 42 ounces, standard deviation (σ) = 10 ounces.

a) 99.7% falls within three standard deviations. Therefore:

99.7% falls within μ ± 3σ = 42 ± 3(10) = 42 ± 30 = (12, 72)

Therefore 99.7% falls within 12 ounce and 72 ounce.

b) For x > 26

$z=\frac{26-42}{10}=-1.6$

For x < 66

$z=\frac{66-42}{10}=2.4$

From the normal distribution table, P(26 < x < 66) = P(-1.6 < z < 2.4) = P(z < 2.4) - P(z < -1.6) = 0.9918 - 0.0548 = 0.937 = 93.7%

c) For x > 34

$z=\frac{34-42}{10}=-0.8$

From the normal distribution table, P(x > 34) = P(z > -0.8) = 1 - P(z < -0.8) = 1 - 0.2119 = 0.7881 = 78.81%

Answer 2

Answer:

Step-by-step explanation:

Given that:

Mean $\mu$ = 42

standard deviation $\sigma$ = 10

Using Empirical Rule:

$\mu$  - $\sigma$ = 42 - 10 = 32                   $\mu$  + $\sigma$ = 42 + 10 = 52                  

$\mu$  - 2$\sigma$ = 42 - 2(10) = 22            $\mu$  + 2$\sigma$ = 42 + 2(10) = 62         

$\mu$  - 3$\sigma$ = 42 - 3(10) = 12             $\mu$  + 3$\sigma$ = 42 + 3(10) = 72      

The curve is attached in the image below.

a). the widget of 99.7% lies between 12 and 72

b) 68 + 13.5 = 81.5%

c) 50 + 34 = 84%