Video Transcript
In a final exam taken by 25
students, 12 students got seven points, six students got eight points, and seven
students got nine points. Given that π denotes the number of
points scored, find the variance of π.
The number of points scored by the
students in this exam is a discrete random variable, which weβre told to denote by
π. We can write down the probability
distribution function of this discrete random variable using the information given
in the question. The values this discrete random
variable can take are the number of points scored by the students, which was either
seven, eight, or nine. The corresponding probabilities,
which we denote using the function π of π₯, are all fractions with denominators of
25, which was the total number of students who took the exam.
12 students got seven points, so
the probability that π₯ equals seven is 12 out of 25. Six students got eight points, so
the probability that π₯ equals eight is six out of 25. And finally, seven students got
nine points, so the probability that π₯ equals nine is seven out of 25. We can confirm that the sum of
these three probabilities is one as it should be for the sum of all the
probabilities in a probability distribution.
Weβve been asked to calculate the
variance of this discrete random variable π, so letβs recall the formula for doing
so. Itβs the expected value of π
squared minus the square of the expected value of π. We need to be clear on the
difference in notation here. For the expected value of π
squared, weβre squaring the variable first and then finding its expectation, whereas
for the second term, weβre calculating the expected value of π and then squaring
this value.
The formulae for calculating each
of these statistics are as follows. The expected value of π is the sum
of each π₯-value multiplied by its corresponding probability. The expected value of π squared is
the sum of each π₯-value squared multiplied by the probability for that
π₯-value. Weβll add some rows to our table to
work the values we need out.
In the first new row, weβll
multiply each π₯-value by its probability. Seven multiplied by 12 over 25 is
84 over 25. Eight multiplied by six over 25 is
48 over 25. And nine multiplied by seven over
25 is 63 over 25. The expected value of π is the sum
of these three values, which is 195 over 25, or 39 over five in simplified form. The next row we add to the table is
for the π₯ squared values, which are 49, 64, and 81. And in the final row of the table,
weβll multiply these values by the corresponding probabilities, giving 588 over 25,
384 over 25, and 567 over 25. The expected value of π squared is
the sum of the three values in the final row of the table, which is 1,539 over
25.
Weβve now calculated the expected
value of π and the expected value of π squared. So weβre ready to substitute these
values into the variance formula. Doing so gives 1,539 over 25 minus
39 over five squared. Evaluating the square gives 1,539
over 25 minus 1,521 over 25, which is 18 over 25. So by first writing down the
probability distribution function of the discrete random variable π, which
represents the number of points scored by the students, weβve found that the
variance of π is 18 over 25.