Question

Mopeds (small motorcycles with an engine capacity below 50 cm^3) 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 46.8 km/h and standard deviation 1.75 km/h. Consider randomly selecting a single such moped.
Required:
a. What is the probability that maximum speed is at most 49 km/h?
b. What is the probability that maximum speed is at least 48 km/h?
c. What is the probability that maximum speed differs from the mean value by at most 2.5 standard deviations?

Answer 1

Answer:

a) 0.1038 = 10.38% probability that maximum speed is at most 49 km/h

b) 0.2451 = 24.51% probability that maximum speed is at least 48 km/h

c) 0.9876 = 98.76% probability that maximum speed differs from the mean value by at most 2.5 standard deviations

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean value 46.8 km/h and standard deviation 1.75 km/h.

This means that $\mu = 46.8, \sigma = 1.75$

a. What is the probability that maximum speed is at most 49 km/h?

This is the pvalue of Z when X = 49. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{49 - 46.8}{1.75}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962

1 - 0.8962 = 0.1038

0.1038 = 10.38% probability that maximum speed is at most 49 km/h.

b. What is the probability that maximum speed is at least 48 km/h?

This is 1 subtracted by the pvalue of Z when X = 48. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{48 - 46.8}{1.75}$

$Z = 0.69$

$Z = 0.69$ has a pvalue of 0.7549

1 - 0.7549 = 0.2451

0.2451 = 24.51% probability that maximum speed is at least 48 km/h.

c. What is the probability that maximum speed differs from the mean value by at most 2.5 standard deviations?

Zscores between Z = -2.5 and Z = 2.5, which is the pvalue of Z = 2.5 subtracted by the pvalue of Z = -2.5

Z = 2.5 has a pvalue of 0.9938

Z = -2.5 has a pvalue of 0.0062

0.9938 - 0.0062 = 0.9876

0.9876 = 98.76% probability that maximum speed differs from the mean value by at most 2.5 standard deviations