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maw [93]
2 years ago
15

9.2 An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a sta

ndard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.
Mathematics
1 answer:
VladimirAG [237]2 years ago
3 0

Answer:

765,795 = 96%

Step-by-step explanation:

confidence interval = 0.04  

The Za/2 theorem = 1/2 = 0.04/2 = 0.02= /x = 720z

If ; 0.02 = 2.05  then the interval is 780-2.05 x 40/√30 x 780+2.05 x 30/√30 = 765,795 = 96%

We see 40/ √30 which is found in equation of finding the sample mean at point /x = 720z

σ 40/ n√30 = 7.3029674334 and is simply a fraction of /x 720z

By normal distribution we find

The 96% confidence interval for the population mean of all bulbs = 765,795

As 765, x 1.04 = 795 = 765, 795

To find Sampling mean.

The Sampling Distribution of the Sample Mean. If repeated random samples of a given size n are taken from a population of values for a quantitative variable, where the population mean is μ (mu) and the population standard deviation is σ (sigma) then the mean of all sample means (x-bars) is population mean μ (mu).

Confidence Level z*-value

80% 1.28

90% 1.645 (by convention)

95% 1.96

96% 2.05

98% 2.33

99% 2.58

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Margo deposited $100 into a savings account earning 4.5% simple annual interest. At the end of each year, she adds $100 to her a
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Answer:

                    principal     interest              total money in account

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Year one       $100             $4.5                              $104.5

year two        $200          $ 9 +$4.5                         $ 213.5

year three      $300          $ 9 +$4.5 + $13.5           $ 327

Step-by-step explanation:

Simple interest for any principal is given by

I = p* r* t/100

I = interest rate accrued on principle amount

p is the amount deposited

r is the rate of interest

t is the time period of saving

_______________________________________________

For year one

p = $100

r = 4.5%

t=1

I = 100*4.5*1/100 = 4.5

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For year two $100 more is added to already existing $100 in account.

p = 100 +100 = $200

r = 4.5%

t=1

I = 200*4.5*1/100 = 9

_______________________________________________

For year two $100 more is added to already existing $200 in account after two years.

p = 100 +100 +100 = $300

r = 4.5%

t=1

I = 300*4.5*1/100 = 13.5

_______________________________________________

There fore total  money in Margo account is

$300 saving deposited by her

$4.5 + $9 + $13.5 = $27 (interest accrued in three time)

Formulating the results in tabular form

                     principal     interest    total money in account

                                                            at the end of year

Year one       100               4.5                      104.5

year two        200           9 +4.5                     213.5

year three      300           9 +4.5 + 13.5           327

7 0
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Alex's friend Pablo comes up with an exact equation to find out how many bags he needs. Use his equation
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Answer:

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Step-by-step explanation:

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Substitute the value of x

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