Question

The scores on a statistics test had a mean of $81$ and a standard deviation of $9$.

One student was absent on the test day and his score wasn't included in the calculation.

If his score of $78$ was added to the distribution of scores, what would happen to the mean and standard deviation

• A. Mean will increase, standard deviation will decrease.
• B. Mean will increase, standard deviation will increase
• C. Mean will decrease, standard deviation will stay the same.
• D. Mean will decrease, standard deviation will increase.
• E. Mean will decrease, standard deviation will decrease.

Answer 1

The correct option is E

Mean will decrease, standard deviation will decrease.

Explanation:

Intent Entity : What changes happen to the mean and standard deviation

Commercial Entity : A statistics test had a mean of $81$ and a standard deviation of $9$.

A student was absent and his expecting score of $78$ is added.

Finding the changes occur in mean and standard deviation :

Let the total number of students be $m$.

Mean = $81$, Standard Deviation = $9$

As we see, the score of the guy who was absent, has lower score than the average score.

As, $\text{Mean}=\frac{\text{Total marks}}{\text{Total number of students}}$

That means, after adding the score of the last student, it will lower the mean of the whole class.

$\text{Standard Deviation}=\sqrt{\frac{1}{n-1}\sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}\left(\text{where,}n=\text{number of students,}{x}_i=\text{individual scores,}\overline{x}=\text{mean score}\right)$

Equating standard deviation,

$\Rightarrow 9=\sqrt{\frac{1}{m-1}\times y}\left(n=m\left(let\right),\sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2=y\right)$

$$\begin{gathered} \Rightarrow 81=\frac{y}{m-1} \\ \Rightarrow y=81m-81\dots \dots \left(1\right) \end{gathered}$$

Now using same formula , after addition of marks of absent student.

Total number of students =$m+1$

Now finding $S.D$

$$\begin{gathered} \Rightarrow NewS.D=\sqrt{\frac{1}{\left(m+1\right)-1}\left(y+{\left(78-81\right)}^2\right)}\left(\text{replacing}n=m+1\right) \\ \Rightarrow NewS.D=\sqrt{\frac{1}{m}\left(y+9\right)} \\ \Rightarrow NewS.D=\sqrt{\frac{81m-81+9}{m}}(\text{putting value of}y\text{from equation}1) \\ \Rightarrow NewS.D=\sqrt{\frac{81m-72}{m}} \end{gathered}$$

$m$ will always a whole number and be greater that the previous number of students.

Hence, we can say $NewS.D.\alpha \frac{1}{Numberofstudents}$.

And the student count is rising so, $S.D$ will decrease.

Hence, Mean will decrease as well as Standard deviation will decrease.

Thus, the correct answer is option $\left(E\right)$.