Question

On an assembly line that fills 8-ounce cans, a can will be rejected if its weight is less than 7.90 ounces. In a large sample, the mean and the standard deviation of the weight of a can is measured to be 8.05 and 0.05 OZ, respectively. (a) Calculate the percentage of the cans that is expected to be rejected on the basis of the given criterion. (b) If the filling equipment is adjusted so that the average weight becomes 8.10 OZ, but the standard deviation remains 0.05 OZ, calculate the rejection rate (% of cans being rejected) . (c) If the filling equipment is adjusted so that the average weight remains 8.05 OZ, but the standard deviation is reduced to 0.03 OZ, calculate the rejection rate.

Answer 1

Answer:

a) 0.001

b) 0.00

c) 0 .00        

Step-by-step explanation:

We are given the following in the question:

Mean = 8.05 oz

Standard deviation = 0.05 oz

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) A can will be rejected if its weight is less than 7.90 ounces

x = 7.90 ounces

P(weight is less than 7.90 ounces)

$P( x < 7.90) = P( z < \displaystyle\frac{7.90 - 8.05}{0.05}) = P(z < -3)$

Calculation the value from standard normal z table, we have,  

$P(x < 7.90) = 0.001$

Only 0.1% of cans will be rejected

b) Mean = 8.10 oz

x = 7.90 ounces

P(weight is less than 7.90 ounces)

$P( x < 7.90) = P( z < \displaystyle\frac{7.90 - 8.10}{0.05}) = P(z < -4)$

Calculation the value from standard normal z table, we have,  

$P(x < 7.90) = 0.00$

Thus, no cans will be rejected.

c) Mean = 8.05 oz

Standard deviation = 0.03 oz

x = 7.90 ounces

P(weight is less than 7.90 ounces)

$P( x < 7.90) = P( z < \displaystyle\frac{7.90 - 8.05}{0.03}) = P(z < -5)$

Calculation the value from standard normal z table, we have,  

$P(x < 7.90) = 0.00$

Thus, no cans will be rejected