Question

Number of tigers in a reserve is normally distributed with mean and variance respectively as $1200and9\times {10}^4$. The probability of finding more than 1800 tigers is approximately.

• A. 0.0125
• B. 0.025
• C. 0.05
• D. None of these

Answer 1

The correct option is B 0.025

Given,
Mean, $\mu =1200$
Variance, ${\sigma }^2=9\times {10}^4$
$\Rightarrow Standarddeviation,\sigma =\sqrt{9\times {10}^4}=300$

Using Standard normal curve,

Probability of finding tigers between

$(\mu -2\sigma )\&(\mu -2\sigma )=0.95$

$\mu -2\sigma =1200-2\times 300=600$

$\mu +2\sigma =1200+2\times 300=1800$

i.e. $P(600\le X\le 1800)=0.95$

$\Rightarrow P(X\le 600)+P(X\ge 1800)=0.05$

Since normal curve is symmetric wrt mean value,
So, $P(X\le 600)=P(X\ge 1800)$
$\Rightarrow 2P(X\ge 1800)=0.05$
$\Rightarrow P(x\ge 1800)=0.025$