Question

The highway mileage (mpg) for a sample of 8 different models of a car company can be found below. Find the mean, median, mode, and standard deviation. Round to one decimal place as needed.19, 22, 26, 27, 30, 32, 35, 35Mean =Median =Mode =Standard Deviation =

Answer 1

The mean, median, mode, and standard deviation for the given sample are 27.6 mpg, 29.5 mpg, 35 mpg, and 4.76 mpg respectively.

Step-by-step explanation:

To find the mean, add up all the values and divide by the number of values:

Mean = (19 + 22 + 26 + 27 + 30 + 32 + 35 + 35) / 8 = 27.6 mpg

To find the median, arrange the values in ascending order and find the middle value:

Median = 29.5 mpg

To find the mode, look for the value(s) that occur(s) most frequently:

Mode = 35 mpg (occurs twice)

To find the standard deviation, first find the variance by subtracting the mean from each value, squaring the differences, finding the mean of the squared differences, and taking the square root:

Variance = [(19-27.6)^2 + (22-27.6)^2 + (26-27.6)^2 + (27-27.6)^2 + (30-27.6)^2 + (32-27.6)^2 + (35-27.6)^2 + (35-27.6)^2] / 8 = 22.65 mpg^2

Standard Deviation = sqrt(22.65) = 4.76 mpg

Answer 2

Sample:

19, 22, 26, 27, 30 , 32, 35, 35

The mean is given by the following formula:

`\mu=(\Sigma X_i)/(n)`

Where n is the size if the sample, in this case n=8.

Replacing the values:

`\mu=(19+22+26+27+30+32+35+35)/(8)=28.25\approx28.3`

The mean is 28.3.

Median:

To find the median of the sample we are going to find the values that are in the middle of the sample, in this case:

27 and 30 are the data. Therefore, the median is:

`Median=(27+30)/(2)=28.5`

The median is: 28.5.

Mode:

The mode is the value that occurs most frequently in the sample, in this case:

The mode is 35.

Standard desviation:

The standar desviation is given by the following formula:

`\sigma=\sqrt[\placeholder{⬚}]{(\Sigma(X_i-\mu)^2)/(n)}`

Where, mu is the mean=28.3, n the size of the sample=8 and Xi each value of the sample. Replacing:

`\sigma=\sqrt{((19-28.3)^2+(22-28.3)^2+(26-28.3)^2+(27-28.3)^2+(30-28.3)^2+(32-28.3)^2+2*(35-28.3)^2)/(8)}`

The standar desviation is: 5.47, approximately=5.5.