Question

The lifetimes of a certain type of light bulbs follow a normal distribution. If approximately 2.5% of the bulbs have lives exceeding 445 hours, and approximately 16% have lives exceeding 393 hours, what are the mean and standard deviation of the lifetimes of this particular type of light bulbs? Round your answer to the nearest integer. Mean = hours Tries 0/5 Standard deviation = hours

Answer 1

Mean = 339 hours

Standard deviation = 54 hours

Explanation:

We are given that approximately 2.5% of the bulbs have lives exceeding 445 hours

So,P(x>445) = 0.025

P(x<445) =1- 0.025 =0.975

z value for 0.975 is 1.96

Formula :
`z=(x-\mu)/(\sigma)`

So,
`1.96=(445-\mu)/(\sigma)`

`1.96 \sigma =445-\mu`--A

Now we are given that approximately 16% have lives exceeding 393 hours,

So,P(x>393) = 0.16

P(x<393) =1- 0.16 =0.84

z value for 0.84 is 1

Formula :
`z=(x-\mu)/(\sigma)`

So,
`1=(393-\mu)/(\sigma)`

`1 \sigma =393-\mu` ---B

Solve A and B

Substitute the value of
`\sigma`from B in A

`1.96 (393 -\mu) =445-\mu`

`770.28 -1.96\mu=445-\mu`

`770.28 -445=1.96\mu-\mu`

`325.28=0.96\mu`

`(325.28)/(0.96)=\mu`

`338.83=\mu`

Substitute the value in B

`\sigma =393-338.83`

`\sigma =54.17`

So, mean = 339 hours

Standard deviation = 54 hours