How to find percentage given mean and standard deviation?

Question

The speed limit on Union street is 40 km/h. Rachel and Joe measured the speed of passing vehicles over a period of time. They found the set of data to be normally distributed with a mean speed of 36 km/h and a standard deviation of 2 km/h. What percentage of the vehicles passed on Union Street at speed greater than 40 km/h?

If μ=36 σ=2
Z=x-μ /σ
so Z=2

Z>2
R(2)=0,02275

So the percentage of the vehicles passed on Union Street at speed greater than 40 km/h is 2,28%

Is the first time that I try to solve this kind of exercise. Is this correct?

Answer 1

You have vehicle speeds $X \sim \mathsf{Norm}(\mu = 36, \sigma = 2)$ and you seek $$P(X > 40) = P\left(\frac{X - \mu}{\sigma} > \frac{40-36}{2}\right) = P(Z > 2) = 0.02275,$$ where $Z$ has the standard normal distribution and the probability can be found using printed normal tables or software.

You do not define what you mean by $R,$ but the numerical answer is OK.

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Note: Using some kinds of statistical software you can skip the 'standardization step' and get the answer directly. In R statistical software, for example, you could use

1 - pnorm(40, 36, 2)
## 0.02275013

In Minitab 16 you can get $P(X \le 40)$ and then subtract from $1.$

MTB > cdf 40;
SUBC> norm 36 2.

Cumulative Distribution Function 

Normal with mean = 36 and standard deviation = 2

 x  P( X ≤ x )
40    0.977250