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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

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Answer:

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
\sigmafrom 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

User Leonid Glanz
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