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If the mean of a normal frequency distribution of 1000 items is 25 and its standard deviation is 2.5, then its maximum ordinate is

1. $\frac{1000}{\sqrt{2 \pi} } e^{-25}$
2. $\frac{1000}{\sqrt{2 \pi} }$
3. $\frac{1000}{\sqrt{2 \pi} } e^{-2.5}$
4. $\frac{400}{\sqrt{2 \pi} }$

Mean u=25

Standard deviation s=2.5

Probability density function f(x)=(e^-((x-u)^2)/(2s^2))/(s(2pi)^0.5)

Maximum value of f(x) is  1/s(2pi)^0.5 at x=u

Hence , maximum value is 1000/s(2pi)^0.5=400/(2pi)^0.5

Option D

How $\frac{1}{{sigma} * \sqrt[2]{2\prod }}$ comes out to be (D) ?
Maximum ordinate = Number of items x max(probability density function)

Max(f(x)):

d/dx(f(x))=-2(x-u)f(x)

for d/dx(f(x))=0

x=u

At which f(x)=1/s(2pi)^0.5
We know that pdf of normally distributed RV $f(x)=\frac{1}{\sigma \sqrt{2\pi }}e^{\frac{-(x-\mu )}{2\sigma ^{2}}}$

Obviously maximum probability will occur at the mean because mean is a point which is closest to all other data set points and therefore the probability of picking up points close to the mean is the most.

Therefore $x=\mu$, therefore our pdf = $f(x)=\frac{1}{\sigma \sqrt{2\pi }}e^{0}$

Therefore max probability = $\frac{1}{2.5\sqrt{2\pi }}$

Therefore for 1000 items = $\frac{1000}{2.5\sqrt{2\pi }}$=$\frac{400}{\sqrt{2\pi }}$
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