Question

The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 47 and a standard deviation of 7. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 47 and 68

Answer 1

`P(-2<Z<2)=95\%`

Explanation:

From question we are told that

Sample mean
`\=x= 47`

Standard deviation
`\sigma =7`

Generally the X -Normal is given as

`Z=(x-\=x)/(\sigma)`

`Z=(x-47)/(9)`

Analyzing the range

`P(47<x<65) = P(0< z<2.00)`

`P(47<x<65) = 95/2`

`P(47<x<65) = 47.5\%`

Mathematically

`Z_1 =(47-47)/(9) =0`

`Z_2 =(65-47)/(9) =2`

Empirical rule shows that

`P(-2<Z<2)=95\%`