Question

6).

a.
The accompanying data represent the miles per gallon of a random sample of cars with a three-cylinder, 1.0 liter engine. (a) Compute the z-score corresponding to the individual who obtained 43.7 miles per gallon. Interpret this result. (b) Determine the quartiles. (c) Compute and interpret the interquartile range, IQR. (d) Determine the lower and upper fences. Are there any outliers? E Click the icon to view the data. (a) Compute the z-score corresponding to the individual who obtained 43.7 miles per gallon. Interpret this result. The 2-score corresponding to the individual is 1 and indicates that the data value is 1 standard deviation(s) above the mean. (Type integers or decimals rounded to two decimal places as MPG Data (b) Determine the quartiles. Q1 = 1 mpg (Type an integer or a decimal. Do not round.) Q2-1 mpg (Type an integer or a decimal. Do not round.) 32.7 34.5 34.8 35.7 35.9 36.3 37.3 37.7 37.9 38.0 38.3 38.5 38.6 38.8 39.5 39.8 39.9 40.7 41.4 41.6 42.5 42.7 43.7 48.8 Q3 = 1 mpg (Type an integer or a decimal. Do not round.) Print Done (c) Compute and interpret the interquartile range, IQR. Select (Type an integer or a decimal. Do not round.) A. The interquartile range is 1 mpg. It is the range of all of the observations in the data set. B. The interquartile range is mpg. It is the range of the middle 50% of the observations in the data set. C. The interquartile range is mpg. It is the range of the observations between the lower and upper fences. OD. The interquartile range is 25% of the observations. mpg. It is the range of the observations between either the lower or upper quartile and the middle quartile; it captures (d) Determine the lower and upper fences. Are there any outliers? The lower fence is 1 (Type an integer or a decimal. Do not round.)
b.

Answer 1

a)Z score=

mean of the MPG=39.025

standard deviation of the MPG=3.49

therefore z score of 43.7 = 1.33

The z score is 1.33 standard deviations avove the mean

b) Since there are even no. of data, the quartiles are

Q1=(n/4 th value + (n/4 +1)th value)/2 = 36.55

Q2=(n/2 th value + (n/2 +1)th value)/2=38.55

Q3=(3n/4 th value + (3n/4 +1)th value)/2=41.375

c)The IQR= Q3-Q1=4.825

The IQR is 4.825. It is the range of the middle 50% observations in the data set

d)The lower fence=Q1-1.5*IQR = 29.31

The upper fence=Q3+1.5*IQR= 48.61

since all the data points are between the lower and upper fences there are no outliers

6b)option C. The distribution is likely to be skewed to the right since the mean is greater than the median. The presence of outliers on the high end could explain the higher standard deviation and the greater mean.