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Lorico [155]
3 years ago
6

The lifetime of a 2-volt non-rechargeable battery in constant use has a Normal distribution with a mean of 516 hours and a stand

ard deviation of 20 hours. Ninety percent of all batteries have a lifetime less than 541.60 hours.

Mathematics
1 answer:
Dmitrij [34]3 years ago
3 0

Answer:

True

Step-by-step explanation:

Data

mean (\mu) = 516 hours

standard deviation (\sigma) = 20 hours

expected lifetime (X) = 541.6 hours

In the figure attached, standard normal distribution table can be seen. Z is computed as follows:

Z = \frac{X - \mu}{\sigma}

Z = \frac{541.6 - 516}{20}

Z = 1.28

In table can be seen that the area between 0 and 1.28 is 0.3997, or simply 0.4. The area until Z = 0 is 0.5; so, the total area until Z = 1.28 is 0.9. That means 90% of batteries would have 541.6 lifetime hours.

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(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

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Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

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Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

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Alternative hypothesis: Sample is significantly different from population

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- the standard t- distribution table

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