Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of operation, and low cost. Suppose the maximum speed of a moped is normally distributed with mean value 46.7 km/h and standard deviation 1.75 km/h. Consider randomly selecting a single such moped. (a) What is the probability that maximum speed is at most 50 km/h? (Round your answer to four decimal places.) (b) What is the probability that maximum speed is at least 47 km/h? (Round your answer to four decimal places.) (c) What is the probability that maximum speed differs from the mean value by at most 1.5 standard deviations? (Round your answer to four decimal places.)

We will calculate the probability based on a random sample of one moped out of the population, normally distributed with mean 46.7 and standard deviation 1.75.

a) This means we have to calculate P(x<50).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{50-46.7}{1.75}=\dfrac{3.7}{1.75}=2.114\\\\\\P(X<50)=P(z<2.114)=0.9827[/tex]

b) We have to calculatee P(x>47).

We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:

[tex]z=\dfrac{X-\mu}{\sigma}=\dfrac{47-46.7}{1.75}=\dfrac{0.3}{1.75}=0.171\\\\\\P(X>47)=P(z>0.171)=0.4321[/tex]

c) If the value differs 1.5 standard deviations from the mean value, we have a z-score of z=1.5

[tex]z=\dfrac{(\mu+1.5\sigma)-\mu}{\sigma}=\dfrac{1.5\sigma}{\sigma}=1.5[/tex]

So the probability that maximum speed differs from the mean value by at most 1.5 standard deviations is P(-1.5<z<1.5):

[tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)=0.9332-0.0668=0.8664[/tex]