SOLUTION: The life expectancy of light bulbs whose lifetimes are normally distributed with a mean life of 750 hours and with a standard deviation of 80 hours. Show or explain how you determi

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Question 1201653: The life expectancy of light bulbs whose lifetimes are normally distributed with a mean life of 750 hours and with a standard deviation of 80 hours. Show or explain how you determine the appropriate z-score and related percentage.
What percent of light bulbs will last longer than 870 hours?
What percent of light bulbs will last between 730 hours and 850 hours?
What percent of light bulbs will last less than 770 hours?
Inverse Normal Distribution: If the quality control program of the company can consistently eliminate the worst 10% of the bulbs manufactured, the manufacturer can safely offer customers a money-back guarantee on all lights that fail before [_____] hours of burning time.
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Thank you!
Answer by math_tutor2020(3838) About Me  (Show Source):
You can put this solution on YOUR website!

Answers:
  1. percentage = 6.681%
  2. percentage = 49.306%
  3. percentage = 59.871%
  4. lifespan = 647 hours
Each value is approximate.
Round each value according to the instructions your teacher provides.

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Explanations:

Problem 1

x = life span measured in hours
This is some positive real number.

The goal is to find P(x > 870) which computes the probability of getting a random light bulb that has a life span longer than 870 hours.
This is equivalent to finding the percentage of light bulbs that last longer than 870 hours.

mu = 750 = population mean lifetime
sigma = 80 = population standard deviation of the lifetimes

Find the z score when x = 870
z = (x-mu)/sigma
z = (870-750)/80
z = 120/80
z = 1.50

The task of finding P(x > 870) is equivalent to P(z > 1.50) for the mu and sigma values mentioned.

I'll be using this Z table
https://www.ztable.net/
A similar table can be found in the back of your stats textbook.
On that link, scroll down the page to see a few examples how to read that table.

Use such a table to find that
P(z < 1.50) = 0.93319
which means
P(z > 1.50) = 1-P(z < 1.50)
P(z > 1.50) = 1-0.93319
P(z > 1.50) = 0.06681
That leads back to
P(x > 870) = 0.06681
This value is approximate.
The same will apply to nearly every other decimal value I mention, with the exception of something like a z score of z = 1.50 (that decimal value is exact).

Roughly 6.681% of the light bulbs will last longer than 870 hours.

You could use a specialized stats calculator such as this one
https://davidmlane.com/normal.html
as an alternative to using a Z table.

This article goes over a few examples of how to calculate normal distribution probabilities on a TI84 calculator.
https://www.statology.org/normal-probabilities-ti-84-calculator/


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Problem 2

mu = 750
sigma = 80

Compute the z score for x = 730
z = (x-mu)/sigma
z = (730-750)/80
z = -20/80
z = -0.25

Repeat for x = 850
z = (x-mu)/sigma
z = (850-750)/80
z = 100/80
z = 1.25

The task of finding P(730 < x < 850) is equivalent to P(-0.25 < z < 1.25)

Use the Z table to find
P(Z < -0.25) = 0.40129
and
P(Z < 1.25) = 0.89435

So,
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-0.25 < Z < 1.25) = P(Z < 1.25) - P(Z < -0.25)
P(-0.25 < Z < 1.25) = 0.89435 - 0.40129
P(-0.25 < Z < 1.25) = 0.49306

This leads back to
P(730 < x < 850) = 0.49306

Roughly 49.306% of the light bulbs will last between 730 hours and 850 hours.

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Problem 3

mu = 750
sigma = 80

Compute the z score for x = 770
z = (x-mu)/sigma
z = (770-750)/80
z = 20/80
z = 0.25

The task of finding P(x < 770) is equivalent to P(z < 0.25)

The Z table says that
P(Z < 0.25) = 0.59871
so,
P(x < 770) = 0.59871


Therefore, roughly 59.871% of the light bulbs will last less than 770 hours.

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Problem 4

The worst 10% is the lowest 10% of life spans.

We need to find a value of k such that
P(Z < k) = 0.10
This value of k is the marker that separates the bottom 10% from the top 90%.

Use a calculator such as this one
https://davidmlane.com/normal.html
or a TI84
https://www.statology.org/inverse-normal-distribution/

If you use the first link, then click on the "value from an area" option.
From there the instructions should be pretty straight forward. Let me know if you have questions.

You should find that P(Z < -1.282) = 0.10 approximately.

Convert the z score z = -1.282 into its equivalent raw x score.
mu = 750
sigma = 80
z = (x-mu)/sigma
-1.282 = (x-750)/80
-1.282*80 = x-750
-102.56 = x-750
x = -102.56+750
x = 647.44

Let's say we round that to the nearest whole number to get 647
We found that P(x < 647) = 0.10 approximately.
About 10% of the light bulbs last less than 647 hours.