State highway patrol has observed that over a particular stretch of highway in t

Question

State highway patrol has observed that over a particular stretch of highway in the country, the speed of vehicles travelling this road has a normal distribution with a mean of 56 miles per hour, and a standard deviation of 4 miles per hour. What percent of drivers on this stretch of road travel between 52 and 64 miles per hour? What percent of drivers on this stretch of road travel less than 48 miles per hour? Calculate the z-score of a driver travelling 51.3 mph. Calculate the z-score of a driver travelling 68.2 mph. Which driver is more likely to be observed, the driver in part (c) or the driver in part (d)?

Explanation / Answer

Solution:

We are given that a variable follows a normal distribution with mean = 56 and SD = 4

Part a

Solution:

We have to find P(52<X<64)

P(52<X<64) = P(X<64) – P(X<52)

The z score formula is given as below:

Z = (X – mean) / SD

Z score for X = 64

Z = (64 – 56) / 4 = 2

P(Z<2) = 0.97725

P(X<64) = 0.97725

Z score for X = 52

Z = (52 – 56)/4 = -4/4 = -1

P(Z<-1) = 0.158655

P(X<52) = 0.158655

P(52<X<64) = P(X<64) – P(X<52)

P(52<X<64) = 0.97725 - 0.158655

P(52<X<64) = 0.818595

Required probability = 0.818595

Part b

Solution:

We have to find P(X<48)

Z = (X – mean) / SD

Z = (48 – 56)/4 = -2

P(X<-2) = P(X<48) = 0.02275

Required probability = 0.02275

Part c

Solution:

Z = (X – mean) / SD

Z score for X = 51.3

Z = (51.3 – 56) / 4

Z = -1.175

Part d

Solution:

Z = (X – mean) / SD

Z score for X = 68.2

Z = (68.2 – 56) / 4

Z = 3.05

Part e

Solution:

The driver in part c is more likely to be observed.

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