Simplify (3^5)^2/3^-2 3^10 3^9 3^8 3^12

The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assum

Vadim26 [7]

Answer:

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762$

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

$Z = \frac{X - \mu}{s}$

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ has a pvalue of 0.9049

X = 25

$Z = \frac{X - \mu}{s}$

$Z = \frac{25 - 26}{0.762}$

$Z = -1.31$

$Z = -1.31$ has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

$Z = \frac{X - \mu}{s}$

$-1.645 = \frac{X - 26}{0.762}$

$X - 26 = -1.645*0.762$

$X = 24.75$

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

$Z = \frac{X - \mu}{s}$

$1.645 = \frac{X - 26}{0.762}$

$X - 26 = 1.645*0.762$

$X = 27.25$

Between $24.75 and $27.25.

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Which of the following is a rational function

Anestetic [448]

B is the only rational answer.

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What are the prime factors of 34725346753856734573457812356273452743654376547654783674337864537644?

NeTakaya

Answer:

3475347396

Step-by-step explanation:

yep

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3 years ago

Write these two statements as a biconditional: If a triangle is equilateral then it is equangular and if a triangle is equiangul

dexar [7]

Answer:

If a triangle is equilateral, then it must also be equiangular and vice versa.

Step-by-step explanation:

A biconditional statement denotes that both a statement AND its reciprocal statement must be true.

Given the two statements:

If a triangle is equilateral, then it must also be equiangular

and

If a triangle is equiangular, then it must also be equilateral.

As both of these statements are always true, we can combine them into one biconditional statement (meaning two conditions are met)

If a triangle is equilateral, then it must also be equiangular and vice versa.

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