Question

Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal quarters will be rejected
A) 1.96%
B) 0.0196%
C) 2.48%
D) 1.62%

Answer 1

Answer:

A) 1.96%

Step-by-step explanation:

You need to know the band between 5.48 g and 5.82 g, the mean is 5.67 g and the standard deviation is .0.070 g, with that information you can calculate the percentage to the left of the mean and the percentage to the right of the mean and add them.

To know its percentage you have to calculate the z score for both values.

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

x is the value being evaluated

μ is the mean

σ is the standard deviation

For 5.48 g:

$z = \frac{5.48-5.67}{0.070} =-2.714$

and is equivalent to 0.0034

For 5.82 g:

$z = \frac{5.82-5.67}{0.070} = 2.143$

and is equivalent to 0.9838

Those numbers mean that the percentage below 5.48 is equivalent to the 0.34% of the values and the percentage below 5.82 is equivalent to 98.38% of the values. Then the percentage rejected is the percentage above 5.82, 100 - 98.38 = 1.62% plus the percentage below 5.48, 0.34%.

Percentage rejected = 1.62% + 0.34% = 1.96%