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Finger [1]
2 years ago
10

Suppose that scores on an exam are normally distributed with mean 80 and standard deviation 5, and that scores are not rounded.

What is the probability that a student scores higher than 85 on the exam
Mathematics
1 answer:
AfilCa [17]2 years ago
7 0

Answer:

3

Step-by-step explanation:

cus it's two 80s and 2 5 so 5 minus 2 is 3

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A laptop producing company also produces laptop batteries, and claims that the batteries
gregori [183]

Answer:

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours. (P-value = 0).

The null and alternative hypothesis are:

H_0: \mu=4\\\\H_a:\mu< 4

Step-by-step explanation:

<em>The question is incomplete: To test this claim a sample or population standard deviation is needed.</em>

<em>We will estimate that the sample standard deviation is 2 hours, and use a t-test to test that claim.</em>

<em> NOTE (after solving): The difference between the sample mean and the mean of the null hypothesis is big enough to reject the null hypothesis, even when we have a sample standard deviation of 3.5 hours, which can be considered bigger than the maximum standard deviation for the sample.</em>

This is a hypothesis test for the population mean.

The claim is that the batteries power the laptops for significantly less than 4 hours.

Then, the null and alternative hypothesis are:

H_0: \mu=4\\\\H_a:\mu< 4

The significance level is 0.05.

The sample has a size n=500.

The sample mean is M=3.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{2}{\sqrt{500}}=0.0894

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{3.5-4}{0.0894}=\dfrac{-0.5}{0.0894}=-5.5902

The degrees of freedom for this sample size are:

df=n-1=500-1=499

This test is a left-tailed test, with 499 degrees of freedom and t=-5.5902, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours.

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Step-by-step explanation:

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Jose got 80% correct 
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