Question

Suppose the mpg rating of cars is normally distributed. Mean = 43 SD = 2

1. P(at least 40 mpg)
2. P(between 30 and 45)
3. The manufacturer wants a rating that improves 90% of existing cars. What is the minimum mpg that will achieve this goal?

Answer 1

Solution :

Given that ,

mean = = 43

standard deviation = = 2

a)P(x 40) = 1 - P(x   40)

= 1 - P[(x - ) / (40 - 43) / 2]

= 1 -  P(z -1.5)   

  Using z table,

= 1 - 0.0668

= 0.9332

b) P(30 < x < 45) = P[(30 - 43)/ 2) < (x - ) /  < (45 - 43) / 2) ]

= P(-6.5 < z < 1.00)

= P(z < 1.00) - P(z < -6.5)

Using z table,

= 0.8413 - 0

= 0.8413

c) Using standard normal table,

P(Z > z) = 90%

= 1 - P(Z < z) = 0.90  

= P(Z < z) = 1 - 0.90

= P(Z < z ) = 0.10

= P(Z < -1.28 ) = 0.10  

z = -1.28

Using z-score formula,

x = z * +

x = -1.28 * 2 + 43

x = 40.44 mpg