We are given a set of 48 measurements. (Your list included 49 data items—I disregarded the .2, since it seemed like an error. You might need to check the complete list.) We are asked to find the standard deviation; then we are asked to calculate the standard deviation using the method of coding.

(a) To find the standard deviation, we first create the required frequency distribution.

X f M

10.0 - 10.4 3 10.2
10.5 - 10.9 2 10.7
11.0 - 11.4 9 11.2
11.5 - 11.9 17 11.7
12.0 - 12.4 8 12.2
12.5 - 12.9 5 12.7
13.0 - 13.4 4 13.2

Here X is the class, f the frequency of the class, and M is the midpoint of the class.

We need the mean to calculate the variance and standard deviation. To calculate the mean, we sum the products of the midpoint and frequency for each class and divide by the total frequency. 48.

Thus `bar(x)=(sum(f*x))/48=11.78bar(3)` (I am treating the 48 values as a sample—if needed you can calculate as a population. For the mean the answers are the same.)

To calculate the standard deviation of the sample, we take the square root of the variance. To calculate the variance, we substitute into the formula

`s^2=(sum(f(M-bar(x))))/47~~.5674` and the standard deviation `s=sqrt(s^2)~~.753`

(b) Using the method of coding: We can somewhat simplify the calculations by performing a linear transformation on the data set. Adding the same number to every member of the data set increases the mean of the set by that number, while multiplying every number multiplies the mean by that same number. Also note that adding the same number to all data entries does not affect the standard deviation or variance, but multiplying by the same number multiplies the standard deviation.

Thus, we can select a nice transformation. I would choose something like y = 10x - 100, where y is the transformed data value, x the original value.

Apply this transformation to the midpoints of the original distribution to get:

Y f

2 3
7 2
12 9
17 17
22 8
27 5
32 4

Now we can calculate the mean and standard deviation of this distribution to find:

`bar(y)=17.8bar(3)` and `s_(y)~~7.532`

To recover the mean and standard deviation of the original data, we note that we used y= 10x - 100.

Thus `bar(y)=10bar(x)-100 ==> bar(x)=(bar(y)+100)/10=11.78bar(3)`

and `s_y=10s_x ==> s_x=(s_y)/10~~.7532`