Question

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 46.6 km/h and standard deviation 1.75 km/h. Consider randomly selecting a single such moped.

Answer 1

a) P ( X ≤ 49 ) = 0.9149

b) P ( X ≥ 48 ) = 0.2119

c) 0.9876

Step-by-step explanation:

Solution:-

- Lets define a random variable X: the maximum speed of a moped.

- The random variable follows normal distribution with the following parameters:

X ~ Norm ( u , σ^2 )

Where,

u = Mean = 46.6 km/h

σ = Standard deviation 1.75 km/h

Hence,

X ~ Norm ( 46.6 , 1.75^2 ).

a) The probability that the maximum speed of mopeds is atmost 49 km/h?

- To evaluate the probability of P ( X ≤ 49 ). We will find the standard Z-score value using the following formula:

`Z-score = (x-u)/(s) \\\\ Z-score = (49-46.6)/(1.75) \\\\Z-score = 1.37142`

- Now use the standard normal tables to determine the required probability of:

P ( Z < 1.37142 ) = 0.9149

Hence,

P ( X ≤ 49 ) = 0.9149

b) The probability that the maximum speed of mopeds is at-least 48 km/h?

- To evaluate the probability of P ( X ≥ 48 ). We will find the standard Z-score value using the following formula:

`Z-score = (x-u)/(s) \\\\ Z-score = (48-46.6)/(1.75) \\\\Z-score = 0.8`

- Now use the standard normal tables to determine the required probability of:

P ( Z ≥ 0.8 ) = 0.2119

Hence,

P ( X ≥ 48 ) = 0.2119

c) The probability that the maximum speed of mopeds differs from mean by at-most 2.5 standard deviation.

- The required probability is the standard error from the mean value "u" of 2.5 standard deviation.

- We don't need to evaluate the test statistics as we are already given the standard error about mean.

- Using, the standard normal Z-score: The required probability is:

P ( -2.5 < Z < 2.5 ) = 2*P ( Z < 2.5 ) - 1

= 2*0.9938 - 1

= 0.9876