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Lady bird [3.3K]

3 years ago

14

The commute time to work in the U.S. Has a bell shaped distribution with a population mean of 24.4 minutes and a population stan

dard deviation of 6.5 minutes. What proportion of the population has a commute time: a) between 11.4 minutes and 37.4 minutes? (1 pt) b) less than 11.4 minutes? (1 pt) c) greater than 37.4 minutes? (1 pt)

1 answer:

Alexxx [7]3 years ago

5 0

Answer:

1. 0.9544

2. 0.0228

3. 0.0228

Step-by-step explanation:

The computation is shown below;

As we know that

At Normal distribution

$P(X < A) = P(Z < \frac{(A - mean)}{standard deviation})$

As per the question, the data provided is as follows

Mean = 24.4 minutes

Standard deviation = 6.5 minutes

Based on the above information

P(11.4 < X < 37.4) = P(X < 37.4) - P(X < 11.4)

= P(Z < (37.4 - 24.4) ÷ 6.5) - P(Z < (11.4 - 24.4) ÷ 6.5)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544

2. P(X < 11.4) = 0.0228

3. P(X ≥ 37.4) = 1 - P(X < 37.4)

= 1 - 0.9772

= 0.0228

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Following are the solution parts for the given question:

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$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

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$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

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Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

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