Question

The mean score of 1000 students for an examination is 34 and standard deviation is 16.
(i) How many candidates can be expected to obtain marks between 30 and 60 assuming the normality of the distribution.
(ii)Determine the limits of the marks of the central 70% of the candidates.
If we are given that {P [0 < z < 0.25] = 0.0987; P [0 < z < 1.63] = 0.4484;P [0 < z < 1.04] = 0.35}

Answer 1

Hint: (i) Firstly, we will find the z – score of candidate who gets P (30 < X < 60) by using the formula \[z=\dfrac{X-\mu }{\sigma }\] and thus find P(${{z}_{1}}$ < z < ${{z}_{2}}$) and then split it up by symmetry and get the number of students. (ii) For this, we will find the values of ${{z}_{1}}$ and ${{z}_{2}}$ from labeled areas and get data points ${{X}_{1}}$ and ${{X}_{2}}$ by using substituting in the formula \[z=\dfrac{X-\mu }{\sigma }\].

Complete step by step answer:
We know that the total number of students who attended the exam = N = 1000.
So, the mean score of 1000 students is 34.
We also know that mean is represented as $\mu $.
So, we get the mean score as $\mu =34$.
We know that standard deviation is represented as \[\sigma \]and we get the value of it as \[\sigma \] = 16.
(i) Now we need to find the number of candidates who obtained marks between 30 and 60. Let X be the required number. So we can write P (30 < X < 60).
Thus let us find z – score = data point – mean / standard deviation.
\[z=\dfrac{X-\mu }{\sigma }....\left( 1 \right)\]
First let us substitute the value as X = 30 in the above expression as:
\[\begin{align}
  & {{z}_{1}}=\dfrac{30-34}{16} \\
 & \Rightarrow {{z}_{1}}=\dfrac{-4}{16} \\
 & \Rightarrow {{z}_{1}}=\dfrac{-1}{4} \\
 & \Rightarrow {{z}_{1}}=-0.25 \\
\end{align}\]
Now, we proceed for the second condition substitute X = 60 in the same expression and we get:
\[\begin{align}
  & {{z}_{2}}=\dfrac{60-34}{16} \\
 & \Rightarrow {{z}_{2}}=\dfrac{26}{16} \\
 & \Rightarrow {{z}_{2}}=\dfrac{13}{8} \\
 & \Rightarrow {{z}_{2}}=1.625 \\
 & \Rightarrow {{z}_{2}}\approx 1.63 \\
\end{align}\]
Thus, \[{{z}_{1}}=-0.25\]and \[{{z}_{2}}=1.63\].
Thus ,with the above values we can write that,
P(${{z}_{1}}$ < z < ${{z}_{2}}$) = P (-0.25 < 0 < 1.63)
P(${{z}_{1}}$ < z < ${{z}_{2}}$) = P (0 < z < 0.25) + P (0 < z < 1.63)
Due to symmetry we can split P (-0.25 < 0 < 1.63) as above.
We have been given the question that P (0 < z < 0.25) = 0.0987 and P (0 < z < 1.63) = 0.4484.
So, we get the value as:
P (-0.25 < 0 < 1.63) = P (0 < z < 0.25) + P (0 < z < 1.63)
$\Rightarrow $P (-0.25 < 0 < 1.63) = 0.0987 + 0.4484
$\Rightarrow $P (-0.25 < 0 < 1.63) = 0.5471
Thus we have the number of students as n whose value is 1000 scoring between 30 and 60 as:
$\begin{align}
  & 0.5471\times n=0.5471\times 1000 \\
 & \Rightarrow 547.1\approx 547 \\
\end{align}$
(ii) Limit of central 70% of candidates.
So, we have values of \[{{z}_{1}}\] from the area label for the area, 0.35 = - 1.04 as, \[{{z}_{1}}\] lies in the left of z = 0.
Similarly, \[{{z}_{2}}\] = 1.04.
So, we get the values as:
\[{{z}_{1}}=\dfrac{{{X}_{1}}-34}{16}\] and \[{{z}_{2}}=\dfrac{{{X}_{2}}-34}{16}\].
Now, by substituting the values of ${{z}_{1}}$ and ${{z}_{2}}$, we get the values as:
\[-1.04=\dfrac{{{X}_{1}}-34}{16}\]and \[1.04=\dfrac{{{X}_{2}}-34}{16}\].
\[{{X}_{1}}=\left( -1.04\times 16 \right)+34\]and\[{{X}_{2}}=\left( 16\times 1.04 \right)+34\]
\[{{X}_{1}}=-16.64+34=17.36\]and \[{{X}_{2}}=16.64+34=50.64\].
Thus we got, \[{{X}_{1}}=17.36\] and \[{{X}_{2}}=50.64\].
Hence, 70% of the candidates score between 17.6 and 50.64.
Thus, the number of candidates who scored between 30 and 60 is 547.70% of the candidates score between 17.36 and 50.64.

Note:
A z –score (z) measures exactly how many standard deviations (\[\sigma \]) above or below the mean data points. A positive z – score says the data point is above average. A negative z – score says that the data point is below average. If zero then the data point is close to average.