The following sample data represent the gasoline mileages (in miles per gallon) determined for cars in a particular weight class:
25.1 29.0 34.5 35.7 37.9 34.9 24.3 26.6 27.3 32.0 30.0 34.5 35.3 33.5 36.6 34.8 16.2 13.1 24.5 33.6 28.0 33.9 30.7 32.0 37.7 21.1 31.2 35.6 34.4 25.2 35.9 18.3 29.4 29.5 34.8 29.4 26.4 38.8 36.0 28.7 23.4 35.3 33.7 38.1 28.6 34.2 34.8 39.2 39.9 36.8
1. using 10 mpg as the lower limit of the first class interval, construct a histogram with intervals of width 5 mpg. what is the midpoint of the cell with the largest number of observations (the mode).
2. is the histogram skewed to the left or right or symmetric?

1) Explanation and figure attached below.

2) For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.

Step-by-step explanation:

Part 1

If we order the data from the smallest to the largest we got:

13.1 , 16.2  ,18.3 , 21.1 , 23.4 , 24.3 , 24.5

25.1 , 25.2 , 26.4 , 26.6 ,27.3  ,28 , 28.6 ,

28.7 , 29 , 29.4 , 29.4 , 29.5 , 30 , 30.7

31.2 , 32 , 32 , 33.5 , 33.6 , 33.7 , 33.9

34.2 , 34.4 , 34.5 , 34.5 , 34.8 , 34.8

34.8 , 34.9 , 35.3 , 35.3 , 35.6 , 35.7

35.9 , 36 , 36.6 , 36.8 , 37.7 , 37.9

38.1 , 38.8 , 39.2 , 39.9

For this case if we use 10 mpg as the lower limit and with a width of 5 mpg for the intervals we have the following table:

Interval   Frequency   Midpoint

[10-15)           1                 12.5

[15-20)          2                17.5

[20,25)         4                 22.5

[25-30)         12               27.5

[30-35)          17              32.5

[35-40)          14              37.5

Total             50

So the histogram is on the figure attached. The midpoint for the class interval with the largest number of observations is 32.5.

Part 2

For this case we see that most of the values are on the right part of the distribution so then we can conclude that the distribution is skewed to the left. The mean seems to be < Median< Mode for this case.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

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step-by-step explanation:

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