The following distribution gives the top speeds (in kilometers per hour) at which thirty races were clocked in auto race. Top Speed 145 150 160 170 No. of racers 9 8 11 2 Find the mean and variance of the distribution of the top speeds?

1 Answer
Feb 24, 2018

Mean: #barx=153.5#.
Variance: #"Var"(x)~~58.583#.
(Or: #hat"Var"(x)~~60.603#.)

Explanation:

The mean #barx# of the top speeds is the sum of all the top speeds, divided by how many there are:

#barx=(sum_(i=1)^n x_i)/n = (x_1+x_2+...+x_n)/n#

There are 9 racers with a top speed of 145.
There are 8 racers with a top speed of 150.
There are 11 racers with a top speed of 160.
There are 2 racers with a top speed of 170.

So,

#barx=(9(145)+8(150)+11(160)+2(170))/(9+8+11+2)#

#color(white)barx=(1305+1200+1760+340)/(30)#

#color(white)barx=4605/30 = 153.5#

The mean is #barx=153.5.#

————

The variance is the average squared difference between each data point and the mean. To calculate the variance, take the difference between each point and the mean, square these differences, add them up, and divide by how many there are:

#"Var"(x)=(sum_(i=1)^n (x_i-barx)^2)/(n)#

#color(white)("Var"(x))=(9(145-153.5)^2+...+2(170-153.5)^2)/(9+8+11+2)#

#color(white)("Var"(x))=(650.25+98+464.75+544.5)/(30)#

#color(white)("Var"(x))=(1757.5)/(30)=58.58bar3#

The variance is #"Var"(x)~~58.583.#

Note: this assumes the given data constitutes a population. Normally, for a calculation like this, these data points represent a sample instead, and so the formula becomes

#hat"Var"(x)=(sum_(i=1)^n (x_i-barx)^2)/(n-1)#

#color(white)(hat"Var"(x))=1757.5/29~~60.603#

The difference here is we are now estimating the population variance from a sample. Dividing by #n-1# instead of #n# makes the estimate unbiased, meaning that the expected value of the estimate is the true population variance.