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

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

& \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{aligned}$$ Therefore, the height below which $$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{aligned} & \Rightarrow B=0.99\times 300 \\\ & \Rightarrow B=198 \\\ \end{aligned}$$ Therefore, we can say 198 students have height less than $$99{\scriptstyle{}^{0}/{}_{0}}$$ that is 72.2 inches.