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Lorico [155]
3 years ago
12

Identical sweaters are on sale in two different stores. The sale price in store A is 30% off the regular price of $25. The sale

price in store B is 40% off the regular price of $30. At which store is the sweater a better buy? Explain.
Mathematics
2 answers:
bekas [8.4K]3 years ago
7 0
30% off 25: multiply 25 by .30 to find 30%: 7.5. Now subtract that from the sales price: 25-7.5=17.5
Do the same for store B: .40•30=12. 30-12=18.
So store A is better by $.50 :) hope this helps :)
mr_godi [17]3 years ago
3 0
Store A is $25.00-30% is 7.50 savings= buying it for 17.50
Store B is $30.00-40% is 12.00 savings = buying it for 18.00
Making store A a better deal by $0.50
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Suppose that a random sample of size 36 is to be selected from a population with mean 43 and standard deviation 6. What is the a
lana66690 [7]

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61.70% approximate probability that X will be more than 0.5 away from the population mean

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 36, \sigma = 6, n = 36, s = \frac{6}{\sqrt{36}} = 1

What is the approximate probability that X will be more than 0.5 away from the population mean?

This is the probability that X is lower than 36-0.5 = 35.5 or higher than 36 + 0.5 = 36.5.

Lower than 35.5

Pvalue of Z when X = 35.5. So

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

By the Central Limit Theorem

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

Z = \frac{35.5 - 36}{1}

Z = -0.5

Z = -0.5 has a pvalue of 0.3085.

30.85% probability that X is lower than 35.5.

Higher than 36.5

1 subtracted by the pvalue of Z when X = 36.5. SO

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

Z = \frac{36.5 - 36}{1}

Z = 0.5

Z = 0.5 has a pvalue of 0.6915.

1 - 0.6915 = 0.3085

30.85% probability that X is higher than 36.5

Lower than 35.5 or higher than 36.5

2*30.85 = 61.70

61.70% approximate probability that X will be more than 0.5 away from the population mean

6 0
3 years ago
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