Question

Weights of a certain model of fully loaded gravel trucks follow a normal distribution with mean µ = 8.6 tons and standard deviation ơ = 0.5 tons. What is the probability that a fully loaded truck of this model is:

a.) More than 7 tons?

b.) Less than 9.6 tons?

c.) Between 8.3 and 8.9 tons?

Answer 1

solution

given that

(a)

P(x > 7 ) = 1 - P(x< 7 )

= 1 - P[(x -) / < (7 -8.6) / 0.5]

= 1 - P(z <-3.2 )

Using z table

= 1 -  0.0007

probability=0.9993

(B)

P(X< 9.6) = P[(X- ) / < (9.6-8.6) /0.5 ]

= P(z <2 )

Using z table

= 0.9772

probability=0.9772

(C)

P(8.3< x < 8.9) = P[(8.3-8.6) /0.5 < (x - ) / < (8.9-8.6) /0.5 )]

= P(-0.6 < Z < 0.6)

= P(Z <0.6 ) - P(Z < -0.6)

Using z table   

= 0.7257-0.2743

probability= 0.4514