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Andreas93 [3]
3 years ago
10

Denise sold 24 more dresses in January than in February and 10 fewer dresses in March than in January. In March she sold 28 dres

ses. How many dresses did Denise sell in February?
Mathematics
1 answer:
Svet_ta [14]3 years ago
8 0
Denise sold 14 dresses in February.
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-8(1+8n)-8(6-4n)=-24
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A manufacturer knows that their items have a normally distributed length, with a mean of 15.4 inches, and standard deviation of
Masteriza [31]

Answer:

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

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 establishes 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.

Mean of 15.4 inches, and standard deviation of 3.5 inches.

This means that \mu = 15.4, \sigma = 3.5

16 items are chosen at random

This means that n = 16, s = \frac{3.5}{\sqrt{16}} = 0.875

What is the probability that their mean length is less than 16.8 inches?

This is the p-value of Z when X = 16.8. So

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

By the Central Limit Theorem

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

Z = \frac{16.8 - 15.4}{0.875}

Z = 1.6

Z = 1.6 has a p-value of 0.9452.

0.9452 = 94.52% probability that their mean length is less than 16.8 inches.

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3 years ago
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