Question

The life expectancy of a typical lightbulb is normally distributed with a mean of 2,100 hours and a standard deviation of 40 hours. What is the probability that a lightbulb will last between 1,965 and 2,165 hours?

Answer 1

@satellite73

Answer 2

@kropot72

Answer 3

@ganeshie8

Answer 4

@camerondoherty

Answer 5

First you need to find the z-scores for 1,965 and 2,165.

Answer 6

How?

Answer 7

\[\large z=\frac{X-\mu}{\sigma}\]

Answer 8

What do the symbols mean again?

Answer 9

I know z is z-score

Answer 10

You should know this. How are you studying statistics?

Answer 11

Hold on

Answer 12

I am sorry, I just don't get this

Answer 13

\[\large \mu=population\ mean\]
\[\large \sigma=population\ standard\ deviation\]

Answer 14

X = random variable

Answer 15

So, The z-score=random variables-2100(population mean)/40(population standard deviation)

Answer 16

What would the random variable be?

Answer 17

There are two random variables. One is 1,965 and the other is 2,165. For the question, two fixed values of the random variable have been chosen.

Answer 18

So we get the z-score for 1,965 as follows:
\[\large z _{1}=\frac{1965-2100}{40}=you\ can\ calculate\]

Answer 19

Hold on

Answer 20

.004

Answer 21

Not really. Lets try the calculation in steps. First the numerator:
1965 - 2100 = ?

Answer 22

.0004

Answer 23

Are you using a calculator?

Answer 24

No, I just forgot to add 3 zero

Answer 25

Not 2

Answer 26

.0004