Mathematics, 25.03.2021 18:50 laylay1548

Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample of 225 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.26 and 0.38

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Answer from: angelynb412

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

37% of the company's orders come from first-time customers.

This means that p = 0.37

A random sample of 225 orders will be used to estimate the proportion of first-time-customers.

This means that n = 225

Mean and standard deviation:

$\mu = p = 0.37$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.37*0.63}{225}} = 0.0322$

What is the probability that the sample proportion is between 0.26 and 0.38?

This is the pvalue of Z when X = 0.38 subtracted by the pvalue of Z when X = 0.26.

X = 0.38

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

By the Central Limit Theorem

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

$Z = \frac{0.38 - 0.37}{0.0322}$

Z = 0.31

Z = 0.31 has a pvalue of 0.6217

X = 0.26

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

$Z = \frac{0.26 - 0.37}{0.0322}$

Z = -3.42

Z = -3.42 has a pvalue of 0.0003

0.6217 - 0.0003 = 0.6214

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

Answer from: Quest

answer: $\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

$\frac{sin40^{\circ}}{x} = \frac{sin60^{\circ}}{12}$

which is the required equation for finding the value of x,

$\frac{x}{sin40^{\circ}}= \frac{12}{sin60^{\circ}}$

$x= 12\times \frac{sin40^{\circ}}{sin60^{\circ}}$

$x= 12\times 0.74222719896$

$x= 8.90672638762\approx 8.967$

Answer from: Quest

answer: 1722

step-by-step explanation:

given: the mass of saturn= $5.685\times10^{26}\ kgs$

the mass of mercury= $3.302\times10^{23}\ kgs$

the number of times the mass of saturn is greater than the mass of mercury

$=\frac{\text{mass of saturn}}{\text{mass of mercury}}\\=\frac{5.685\times10^{26}}{3.302\times10^{23}}\\\\=1.72168\times10^{26-23}{by using law of exponent}]\\=1.72168\times10^3=1721.68\approx1722$

hence, the mass of saturn is 1722 times greater than the mass of mercury.

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