Question

If the heights of 300 students are normally distributed with mean 64.5 inches and standard deviation of 3.3 inches. Find the height below which \[99{\scriptstyle{}^{0}/{}_{0}}\] of students lie.

Answer 1

Hint: For solving this problem we use the concept of normal distribution which includes Z – scores. For a normal distribution we take \[Z=\dfrac{X-M}{\sigma }\] where \[M\] is mean, \[\sigma \] is mean deviation and \[X\] is some height corresponding to \[Z\] . We need to find the height below \[99{\scriptstyle{}^{0}/{}_{0}}\] of students which can be considered as 0.99 probability. We have one standard table of values of \[Z\] corresponding to probabilities. By using the table we find the value of \[X\] for corresponding \[Z\] .

Probability	Z
0.80	2.41
0.82	2.40
0.84	2.39
0.87	2.38
0.89	2.37
0.91	2.36
0.94	2.35
0.96	2.34
0.99	2.33
1.0	2.32

Complete step-by-step answer:
We are given that the mean of data is 64.5 inches and mean deviation is 3.3 inches.
Let us assume that the values of mean and mean deviations as
 \[\sigma =3.3\]
 \[M=64.5\]
We are asked to find height below which \[99{\scriptstyle{}^{0}/{}_{0}}\] of students lie.
So, the probability is 0.99. So, for some height \[{{X}_{1}}\] having 0.99 probability, we can write
 \[P\left( X<{{X}_{1}} \right)=0.99\]
We know that for normal distributed data, we take the formula of Z – scores as \[Z=\dfrac{X-M}{\sigma }\] where \[M\] is mean, \[\sigma \] is mean deviation and \[X\] is some height corresponding to \[Z\] .
For some height \[{{X}_{1}}\] having 0.99 probability, we can write
 \[\Rightarrow Z=\dfrac{{{X}_{1}}-M}{\sigma }.....equation(i)\]
We know that from the table the value of \[Z\] corresponding to 0.99 probability is 2.33.
By substituting the required values in equation (i) we get
 \[\begin{align}
  & \Rightarrow 2.33=\dfrac{{{X}_{1}}-64.5}{3.3} \\
 & \Rightarrow {{X}_{1}}=64.5+\left( 2.33\times 3.3 \right) \\
 & \Rightarrow {{X}_{1}}=72.189\simeq 72.2 \\
\end{align}\]
Therefore, the height below which \[99{\scriptstyle{}^{0}/{}_{0}}\] of students lies is 72.2 inches.

Note: Students may make mistakes by not remembering the table of Z – scores. It is the standard table which needs to be remembered for normal distribution of data.
There may be a continuation question to the above question that is to find the number of students lying below \[99{\scriptstyle{}^{0}/{}_{0}}\] . To find the number of students we use the formula
 \[B=P\times N\]
Here, \[B\] is number of students lie below \[99{\scriptstyle{}^{0}/{}_{0}}\]
 \[P\] is the probability given as 0.99
 \[N\] is the total number of students given as 300.
By substituting these values in above formula we get
 \[\begin{align}
  & \Rightarrow B=0.99\times 300 \\
 & \Rightarrow B=198 \\
\end{align}\]
Therefore, we can say 198 students have height less than \[99{\scriptstyle{}^{0}/{}_{0}}\] that is 72.2 inches.