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2 years ago

Find and compare the growth rates in the weekly number of specialty items produced at each factory.

Mathematics

1 answer:

DerKrebs [107]2 years ago

7 0

0.129 is the change in the growth rate of company.

According to statement

we have to find and compare the growth rate of the company weekly.

we use the growth rate formula is

GROWTH RATE = present - past / present

Substitute the values in this formula

1st week = 220-190 / 220

1st week = 0.136

2nd week = 252-220 / 252

2nd week = 32 / 252

2nd week = 0.126

3rd week = 290-252 / 290

3rd week = 0.131

4th week = 337 - 290 / 337

4th week = 0.139

5th week = 380 - 337 / 380

5th week = 0.113

6th week = 440 - 380 / 440

6th week = 0.136

7th week = 505 - 440 / 505

7th week = 0.128

Now we find the average value of growth rate

Average = 0.136 + 0.126+0.131+0.139+0.113+0.136+0.128 /7

Average = 0.909/7

Average = 0.129

So, 0.129 is the change in the growth rate of company.

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Answer:

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Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

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Problems of normally distributed samples 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.

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Normally distributed with mean 375 minutes and standard deviation 68 minutes. So $\mu = 375, \sigma = 68$

What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?

So $n = 6, s = \frac{68}{\sqrt{6}} = 27.76$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 375}{27.76}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.

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Normally distributed with mean 522 minutes and standard deviation 106 minutes. So $\mu = 522, \sigma = 106$