Find the domain of the equations. do not solve

$819.20 is the answer considering each time the units increase by fours, it adds $204.80

8 0

3 years ago

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.

3 0

3 years ago

Help!!! I just need Help on number 6 please!!!!!!!! ASAP!!!

PtichkaEL [24]

Answer:

$3$

Step-by-step explanation:

$\sqrt{12} =3.46$ (2 decimal places)

If you want to arrange the tiles into a square, you will need 3.46 in each row/column.

This is physically impossible, hence you round it down to 3.

(If you round it up, you won't have enough tiles since $4*4=16$ )

Thus, the square area will be $3*3=9$

Tiles left: $12-9=3$

6 0

3 years ago

Given an int variable n that has already been declared and initialized to a positive value, and another int variable j that has

Inessa05 [86]

Answer:

for(j = n; j > 0; j--)

System.out.print(\"*\");

Step-by-step explanation:

-A for loop is a repetition control structure.

-It allows the efficiency to write a loop that would otherwise be written a couple of times.

-The output is a single line comprising n asterisks,

-The number of asterisks printed will be equivalent to the n-value declared.

7 0

2 years ago

One garden had 5 times as many raspberry bushes as a second garden. After 22 bushes are transplanted from the first garden to th

Aleonysh [2.5K]

Let the number of raspberry bushes in one garden = x

And the number of raspberry bushes in second garden = y

Garden one has 5 times as many raspberry bushes as second garden,

So the equation will be,

x = 5y -------(1)

If 22 bushes were transplanted from garden one to the second, number of bushes in both the garden becomes same,

Therefore, (x - 22) = (y + 22)

x - y = 22 + 22

x - y = 44 ------(2)

Substitute the value of x from equation (1) to equation (2)

5y - y = 44

4y = 44

y = 11

Substitute the value of 'y' in equation (1),

x = 5(11)

x = 55

Therefore, Number of bushes in garden one were 55 and in second garden 11 originally.

Learn more,

brainly.com/question/12422372

5 0

3 years ago

Find the domain of the equations. do not solve​

Find the domain of the equations. do not solve