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3 years ago

The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every

Mathematics

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

X ≥ 112,6 $

Step-by-step explanation:

Normal Distribution N ( 95 , 16 )

μ₀ = 95

σ = 16

We need to get z (score) for area 0,1357; then from z - Table we get

z (score) = 1,1

Now z (score ) = ( X - μ₀ )/σ

1,1 = ( X - 95 ) / 16

17,60 + 95 = X

X = 112,6 $

That means that in order to get a gift customes has to spend at least 112,6 $

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Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,

umka21 [38]

It's hard to tell where one set ends and the next starts. I think it's

A. 25, 36, 44, 51, 62, 77

B. 3, 3, 3, 7, 9, 9, 10, 14

C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39

D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.

A. 25, 36, 44, 51, 62, 77

That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.

B. 3, 3, 3, 7, 9, 9, 10, 14

Average around 7, sigma around 4, within 2 sigma, seems ok.

C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39

Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.

D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82

Average around 74, sigma 8, seems very tight.

I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.

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4 years ago

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Once a month, volunteers from the Green Sea Club clean up litter at the beach. Last month, the volunteers picked up 8 pounds of

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The unit rate here is 2 pounds of trash in an hour. If two pounds of trash are picked up in an hour, we can multiply that by 3 and get that there will be 6 pounds picked up in 3 hours.

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Graph the inverse variation

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4 years ago

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

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3 years ago