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Neporo4naja [7]2 years ago

8 0

The probability that a single can will contain 11.6 ounces or less of soda is 0.2843

<h3>Probability that a can contains 11.6 ounces or less</h3>

The given parameters are:

x = 11.6

Mean = 12

Standard deviation = 0.7

Calculate the z value using:

$z = \frac{x - \bar x}{\sigma}$

This gives

$z = \frac{11.6-12}{0.7}$

z = -0.57

The probability is then calculated as:

P(x ≤ 11.6) = P(z ≤ -0.57)

Using the z table of probabilities, we have:

P(x ≤ 11.6) = 0.2843

<h3>Probability that a pack contains 11.6 ounces or less</h3>

In (a), the probability that a can contains 11.6 ounces or less is 0.2843

The probability that all cans in a pack contains 11.6 ounces or less is

P(6) = 0.2843^6

P(6) = 0.00053

<h3>Probability that a case contains 11.6 ounces or less</h3>

In (a), the probability that a can contains 11.6 ounces or less is 0.2843

The probability that all cans in a case contains 11.6 ounces or less is

P(36) = 0.2843^36

P(36) ≈ 0

<h3>Draw three normal distributions</h3>

See attachment for the normal distributions

<h3>The happening on the graph</h3>

The summary of the graph is that, as the sample size increases the probability decreases

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Answer:

Look below.

Step-by-step explanation:

Good ol' Pythagorean Theorem.

To find the short side of the larger triangle, we'll use this equation:

$a^{2} +b^{2} =c^{2} \\\\9^2+b^2=10^2\\81+b^2=100\\$

You can solve the rest on your own.

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3 years ago

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3 years ago

10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the

Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

$H_0: \mu = 0$

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

$H_1: \mu > 0$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

In this problem, the parameters are given as follows:

$\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50$ .

Hence, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}$

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by $t^{\ast} = 2.4$ .

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

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2 years ago

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Answer:

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