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

Answer:

Mean = 339 hours

Standard deviation = 54 hours

Step-by-step 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 : $z=\frac{x-\mu}{\sigma}$

So, $1.96=\frac{445-\mu}{\sigma}$

$1.96 \sigma =445-\mu$--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 : $z=\frac{x-\mu}{\sigma}$

So, $1=\frac{393-\mu}{\sigma}$

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

Solve A and B

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

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

$770.28 -1.96\mu=445-\mu$

$770.28 -445=1.96\mu-\mu$

$325.28=0.96\mu$

$\frac{325.28}{0.96}=\mu$

$338.83=\mu$

Substitute the value in B

$\sigma =393-338.83$

$\sigma =54.17$

So, mean = 339 hours

Standard deviation = 54 hours