Aset of data contains 53 observations. the minimum value is 42 and the maximum value is 129. the data are to be organized into a frequency distribution. (a) how many classes would you suggest? (b) what would you suggest as the lower limit of the first class? (round your answer to the nearest whole number.)

solution;

a)From the given information,

N = 53,min=42 and max = 129

K = 1+3.322log₁₀53

   = 1+3.322(1.724)

    = 1+5.728

    = 6.728

Number of classes is 7.

b)  find the width of the class interval first  

width = max – min /classes  

         = 129-42/7

  = 87/7 =12.42

Take the next largest number 13 as class width. Since the minimum value is 42

The lower limit iof the first class could be 40.

So,

The upper limit of the last class is,

40+(13 x 7) = 131

As the upper limit of the last class covers the entire range of values,

It is better to take the lower limit of the first class as 40

The lower limit of the first calss could be 40.

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.