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Cloud [144]
3 years ago
7

Suppose you are told that a 95% confidence interval for the average price of a gallon of regular gasoline in your state is from

$3.05 to $3.92. Use the fact that the confidence interval for the mean is in the form x − E to x + E to compute the sample mean and the maximal margin of error E. (Round your answers to two decimal places.)
Mathematics
1 answer:
Cloud [144]3 years ago
8 0

Answer:

The sample mean is 3.48 and the margin of error is 0.44.

Step-by-step explanation:

We know that the confidence interval is written in the form "x − E to x + E", meaning x: sample mean or estimator, and E: margin of error.

To calculate the mean, we can calculate the average of both bounds of the confidence interval:

X=(UL-LL)/2=(3.92+3.05)/2=6.97/2=3.48

With one of the bounds, we can calculate the margin of error E

UL=X+E\\\\E=UL-X=3.92-3.48=0.44

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3 years ago
Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

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2 years ago
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(2x² + y²)⁴ = 16x⁸ + 32x⁶y² + 24x⁴y⁴ + 8x²y⁶ + y⁸


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What combination of transformations is shown below?
muminat

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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