Mathematics, 09.03.2021 05:20 lilpeepxliltracy

On any given day 34% of sales at Ruby's jewelry store are from necklaces, while 28% of sales at Nugget Jewels are from necklaces. Suppose Ruby's has 50 customers and Nugget Jewels has 60 customers on a randomly

selected day. Let R = the proportion of sales that are from necklaces at Ruby's and N = the proportion of sales that

are from necklaces at Nugget Jewels. What is the probability that the proportion of sales from necklaces at Ruby's

will be less than the proportion of sales from necklaces at Nugget Jewels?

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Answers

0.2483 = 24.83% probability that the proportion of sales from necklaces at Ruby's will be less than the proportion of sales from necklaces at Nugget Jewels.

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 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.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

Sampling distribution for the sample proportion of necklaces from Ruby's jewelry:

34% of sales at Ruby's jewelry store are from necklaces, 50 customers.

This means that p = 0.34, n = 50

$\mu_{R} = p = 0.34$

$s_{R} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.34*0.66}{50}} = 0.067$

Sampling distribution for the sample proportion of necklaces from Nugget jewels:

28% of sales at Nugget Jewels, 60 customers.

This means that p = 0.28, n = 60

$\mu_{N} = p = 0.28$

$s_{N} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.28*0.72}{60}} = 0.058$

What is the probability that the proportion of sales from necklaces at Ruby's will be less than the proportion of sales from necklaces at Nugget Jewels?

This is P(X < 0) , on the subtraction distribution R - N.

We have that:

$\mu_{R-N} = \mu_{R} - \mu_{N} = 0.34 - 0.28 = 0.06$

$s_{R-N} = \sqrt{s_{R}^2 + s_{N}^2} = \sqrt{0.067^2 + 0.058^2} = 0.0886$

The probability is the pvalue of Z when X = 0. So

$Z = \frac{X - \mu_{R-N}}{s_{R-N}}$

$Z = \frac{0 - 0.06}{0.0886}$

Z = -0.68

Z = -0.68 has a pvalue of 0.2483.

0.2483 = 24.83% probability that the proportion of sales from necklaces at Ruby's will be less than the proportion of sales from necklaces at Nugget Jewels.

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