Question

The speed of the cars on a motorway is normally distributed with mean 90Km/hr and standard deviation 10 km/hr. Determine the probability that a randomly chosen car will have a speed greater than 100 km/hr
[a] 0.65866
[b] 0.7392
[c] 0.2120
[d] 0.52545

Answer 1

Hint: Use the fact that if X is normal distributed random variable with mean $\mu $ and standard deviation $\sigma $, then $P\left( X\le x \right)=\int_{-\infty }^{x}{\dfrac{1}{\sqrt{2\pi }\sigma }{{e}^{-\dfrac{1}{2}{{\left( \dfrac{t-\mu }{\sigma } \right)}^{2}}}}dt}$. Assume that X is a random variable describing the speed of the cars on the motorway. Using the above property determines that a randomly chosen car will be travelling at more than 100 km/hr.

Complete step-by-step answer:
Let X be the random variable describing the speed of the cars on the motorway.
Hence according to question, we have
$X\sim N\left( 90,100 \right)$
We know that $P\left( X\ge x \right)=1-P\left( X\le x \right)$(X is a continuous random variable)
Hence we have
$P\left( X\ge 100 \right)=1-P\left( X\le 100 \right)$
Now, we know that if $X\sim N\left( \mu ,{{\sigma }^{2}} \right)$, then $P\left( X\le x \right)=\int_{-\infty }^{x}{\dfrac{1}{\sqrt{2\pi }\sigma }{{e}^{\dfrac{-1}{2}{{\left( \dfrac{t-\mu }{\sigma } \right)}^{2}}}}dt}$
Hene, we have
$P\left( X\ge 100 \right)=1-P\left( X\le 100 \right)=1-\int_{-\infty }^{100}{\dfrac{1}{\sqrt{2\pi }\left( 10 \right)}{{e}^{\dfrac{-1}{2}{{\left( \dfrac{t-90}{10} \right)}^{2}}}}dt}$
Put $\dfrac{t-90}{10}$ = z.
Differentiating both sides with respect to t, we get
$dz=\dfrac{dt}{10}\Rightarrow dt=10dz$
Hence, we have
$\begin{align}
  & P\left( X\ge 100 \right)=1-\int_{-\infty }^{1}{\dfrac{1}{\sqrt{2\pi }\left( 10 \right)}{{e}^{-\dfrac{1}{2}{{z}^{2}}}}\left( 10dz \right)} \\
 & =1-\int_{-\infty }^{1}{\dfrac{1}{\sqrt{2\pi }}{{e}^{-\dfrac{1}{2}{{z}^{2}}}}dz} \\
\end{align}$
We know that
$\int_{-\infty }^{x}{\dfrac{1}{\sqrt{2\pi }}{{e}^{-\dfrac{{{t}^{2}}}{2}}}dt}=\phi \left( x \right)$
Hence, we have
$P\left( X\ge 100 \right)=1-\phi \left( 1 \right)$
From phi-z table, we have
$\phi \left( 1 \right)=0.34134$
Hence, we have
$P\left( X\ge 100 \right)=1-0.34134=0.65866$
Hence option [a] is correct.

Note: Alternatively, we can use the fact that if X is normally distributed with mean $\mu $ and standard deviation $\sigma $, then the random variable $\dfrac{X-\mu }{\sigma }$ is normally distributed with mean 0 and standard deviation 1.
Hence, we have
$P\left( X\ge 100 \right)=1-P\left( X\le 100 \right)=1-P\left( \dfrac{X-90}{10}\le 1 \right)=1-\phi \left( 1 \right)$, which is the same as obtained above.
Hence option [a] is correct.