Answer:
Variance: 10.9 or 10.7 (see explanation below)
Standard deviation: 3.3
Step-by-step explanation:
Since I do not have the data you fed into your calculator, I am not sure how you got the values you did for variance and standard deviation
This is how I worked out the problem:
Whenever you have equal class intervals and a frequency distribution use the midpoint of each class as the data point
So for the class interval 10-12, the midpoint is (10 + 12)/2 = 11
If we refer to a midpoint as [tex]M_i[/tex] and the frequency as [tex]f_i[/tex] for that midpoint then
[tex]\texttt{Mean }= \mathtt{\dfrac{\sum(M_i\cdot f_i)}{\sum{f_i}}} = \mathtt{\bar{X}}[/tex]
Let [tex]N = \Sigma{f_i}[/tex]
Population Variance is given by
[tex]\sigma^2 = \Sigma\dfrac{f(M_i-\bar{X})^2}{N}[/tex]
Sample Variance is given by
[tex]\sigma^2 = \Sigma\dfrac{f(M_i-\bar{X})^2}{N-1}[/tex]
And standard deviation is given by [tex]\sigma[/tex] = [tex]\sqrt{\sigma^2}[/tex]
I computed the values using two different calculators and came up with two slightly different values for the variance, 10.9 and 10.7.
The standard deviation being the square root and rounded to 1 decimal point is the same at 3.3
My guess is that one of the calculators calculated sample variance and the other the population variance but not sure.
I have provided both sets of values. Try both and see what happens
Sorry, couldn't give a more definitive answer
Hope that works for you
