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Question: The speed of the cars on a motorway is normally distributed with mean 90Km/hr and standard deviation...

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

Explanation

Solution

Hint: Use the fact that if X is normal distributed random variable with mean μ\mu and standard deviation σ\sigma , then P(X≤x)=∫−∞x12πσe−12(t−μσ)2dtP\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∼N(90,100)X\sim N\left( 90,100 \right)
We know that P(X≥x)=1−P(X≤x)P\left( X\ge x \right)=1-P\left( X\le x \right)(X is a continuous random variable)
Hence we have
P(X≥100)=1−P(X≤100)P\left( X\ge 100 \right)=1-P\left( X\le 100 \right)
Now, we know that if X∼N(μ,σ2)X\sim N\left( \mu ,{{\sigma }^{2}} \right), then P(X≤x)=∫−∞x12πσe−12(t−μσ)2dtP\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(X≥100)=1−P(X≤100)=1−∫−∞10012π(10)e−12(t−9010)2dtP\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 t−9010\dfrac{t-90}{10} = z.
Differentiating both sides with respect to t, we get
dz=dt10⇒dt=10dzdz=\dfrac{dt}{10}\Rightarrow dt=10dz
Hence, we have
P(X≥100)=1−∫−∞112π(10)e−12z2(10dz) =1−∫−∞112πe−12z2dz \begin{aligned} & 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{aligned}
We know that
∫−∞x12πe−t22dt=ϕ(x)\int_{-\infty }^{x}{\dfrac{1}{\sqrt{2\pi }}{{e}^{-\dfrac{{{t}^{2}}}{2}}}dt}=\phi \left( x \right)
Hence, we have
P(X≥100)=1−ϕ(1)P\left( X\ge 100 \right)=1-\phi \left( 1 \right)
From phi-z table, we have
ϕ(1)=0.34134\phi \left( 1 \right)=0.34134
Hence, we have
P(X≥100)=1−0.34134=0.65866P\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 μ\mu and standard deviation σ\sigma , then the random variable X−μσ\dfrac{X-\mu }{\sigma } is normally distributed with mean 0 and standard deviation 1.
Hence, we have
P(X≥100)=1−P(X≤100)=1−P(X−9010≤1)=1−ϕ(1)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.