Question

Suppose the speeds of cars along a stretch of I-40 is normally distributed with a mean of 70 mph and standard deviation of 8 mph.

What would be the speed for a car whose z-score is z=−1.6?

f a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed below 70 mph?

If a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed greater than or equal to 86 mph? (Give your answer to 2 decimal

If a car is chosen at random on this stretch of highway, what is the probability it will be travelling at a speed between 70 mph and 86 mph? (Give your answer to 2 decimal

What is the probability that a car chosen at random on this highway will be traveling between 65 mph and 80 mph? (Give your answer to 2 decimal places.)

Answer 1

Solution:-

Given that,

mean = = 70

standard deviation = = 8

a) z = -1.6

Using z-score formula,

x = z * +

x = -1.6 * 8 + 70

x = 57.2 mph

b) P(x < 70) = P[(x - ) / < (70 - 70) / 8]

= P(z < 0)

Using z table,

= 0.50

c) P(x 86) = 1 - P(x   86)

= 1 - P[(x - ) / ( 86 - 70) / 8]

= 1 -  P(z 2.0)   

  Using z table,

= 1 - 0.98

= 0.02

d) P(70 < x < 86) = P[(70 - 70)/ 8) < (x - ) /  < (86 - 70) / 8 ) ]

= P(0 < z < 2.0)

= P(z < 2.0) - P(z < 0)

Using z table,

= 0.98 - 0.50

= 0.48

e) P(65 < x < 80) = P[(65 - 70)/ 8) < (x - ) /  < (80 - 70) / 8 ) ]

= P(-0.63 < z < 1.25)

= P(z < 1.25) - P(z < -0.63)

Using z table,

= 0.89 - 0.26

= 0.63