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3 years ago

HELPPPPPPPPPPPPIIDISKDKDJDNFN

Mathematics

2 answers:

Kisachek [45]3 years ago

6 0

Both are equal to each other

stepladder [879]3 years ago

5 0

$\displaystyle \bf\\\frac{1}{4} \cdot \frac{3}{4} = \frac{3}{4} \cdot \frac{1}{4}~~~\text{Equal because the multiplication is commutative.}\\\\ \\\frac{3}{4}\div\frac{2}{5} \neq \frac{2}{5}\div\frac{3}{4}~~~\text{Not equal because division is not commutative.}$

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2 years ago

The amount people pay for cable service varies quite a bit but the mean monthly fee is $142 and the standard deviation is $29. t

zhuklara [117]

Answer:

a) By the Central Limit Theorem, the mean is $142 and the standard deviation is $0.7488.

b) By the Central Limit Theorem, approximately normal.

c) 0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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.

The mean monthly fee is $142 and the standard deviation is $29.

This means that $\mu = 142, \sigma = 29$

Part a: what are the mean an standard deviation of the sample distribution of x hat show your work and justify your reasoning.

Sample of 1500(larger than 30).

By the Central Limit Theorem

The mean is $142

The standard deviation is $s = \frac{29}{\sqrt{1500}} = 0.7488$

Part b: what is the shape of the sampling distribution of x hat justify your answer.

By the Central Limit Theorem, approximately normal.

Part C: what is the probability that the average cable service paid by the sample of cable service customers will exceed $143?

This is 1 subtracted by the pvalue of Z when X = 143. So

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

By the Central Limit Theorem

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

$Z = \frac{143 - 142}{0.7488}$

$Z = 1.34$

$Z = 1.34$ has a pvalue of 0.9099

1 - 0.9099 = 0.0901

0.0901 = 9.01% probability that the average cable service paid by the sample of cable service customers will exceed $143

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