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jeyben [28]
2 years ago
14

Find the standard deviation of the following data. Answers are rounded to the nearest tenth. 5, 5, 6, 12, 13, 26, 37, 49, 51, 56

, 56, 84
Mathematics
1 answer:
kogti [31]2 years ago
6 0

Answer:

24.2(to the nearest tenth)

Step-by-step explanation:

The question is ungrouped data type

standard deviation =√ [∑ (x-μ)² / n]

mean (μ)=∑x/n

           = \frac{5+5+6+12+13+26+37+49+51+56+56+84}{12}

          =33.3

x-μ   for data 5, 5, 6, 12, 13, 26, 37, 49, 51, 56, 56, 84 will be

                   -28.3, -28.3, -27.3, -21.3, -7.3, 3.7, 15.7,17.7, 22.7,22.7,50.7

(x-μ)² will be 800.89, 800.89,745.29,453.69,53.29,13.69,246.49,313.29,515.29,515.29,2570.49

∑ (x-μ)² will be = 7028.59

standard deviation = √(7028.59 / 12)

                      =24.2

     

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Step-by-step explanation:

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A group of neighborhood children have a lemonade stand. They spend $5 on supplies. The function graphed shows the earnings they
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7 0
1 year ago
In a certain month, Oscar sells 6 cars more than Jackie. If together they sell 30 cars, how many cars did Jackie sell?
diamong [38]

Answer:

Jackie sold 12 cars.

Step-by-step explanation:

If we call the number of cars Oscar sold O, and the number of cars Jackie sold J, we can say the following:

O = J + 6

As Oscar sold 6 cars more than Jackie.

Together, they sold 30 cars.

O + J = 30

Since we know that:

O = J + 6

... we can put this into our previous equation.

O + J = 30

(J + 6) + J = 30

J + J + 6 = 30

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Subtract 6 from both sides:

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Divide both sides by 2:

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8 0
2 years ago
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In 1998 the average income for middle class families in the US was $37,100 with a population standard deviation of $6362. We wan
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Answer:

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

The best answer would be:

A. Z test

Step-by-step explanation:

Data given and notation  

\bar X=36670 represent the mean average

\sigma=6362 represent the population standard deviation for the sample  

n=1225 sample size  

\mu_o =37100 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is equal or not to 37100, the system of hypothesis would be:  

Null hypothesis:\mu =37100  

Alternative hypothesis:\mu \neq 37100  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

The best answer would be:

A. Z test

5 0
2 years ago
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