Question

In a class of 100, the mean on a certain exam was 50, the standard deviation is 0. This means
A. half the class had scores less than 50
B. There was a high correlation between ability and grade.
C. everyone had a score of exactly 50.
D. half the class had 0’s and half had 50’s

Answer 1

Hint: We first find the general form of deviations of n observations and the form of mean and standard deviation for those n observations. We find the formula for the mean using the values of number of observations and the deviation value. We place the values and get the solution.

Complete step by step answer:
We take the weight of the students as ${{x}_{i}}$. The mean of the set of the observations will be \[\overline{x}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\].
It’s given that the mean on a certain exam was 50. Putting the values, we get \[50=\overline{x}=\dfrac{1}{100}\sum\limits_{i=1}^{100}{{{x}_{i}}}\].
Sample standard deviation can be expressed as \[\sigma =\sqrt{{{\sigma }^{2}}}=\sqrt{\dfrac{1}{n}\sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}\].
It is given that the standard deviation is 0. Putting the values, we get \[0=\sqrt{\dfrac{1}{100}\sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}\].
The simplification gives \[0=\sqrt{\dfrac{1}{100}\sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}\Rightarrow \sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}=0\].
The sum of squares is 0 which gives individual terms to be equal to 0.
So, \[\left( {{x}_{i}}-\overline{x} \right)=0\]. The simplification gives \[{{x}_{i}}=\overline{x}\].This means every entry of ${{x}_{i}}=\overline{x}=50$. Everyone had a score of exactly 50.

So, the correct answer is “Option C”.

Note: These all-required attributes are the measurement of central tendency and dispersion. The sample mean is part of central tendency and sample standard deviation is part of dispersion.