Question

CNNBC recently reported that the mean annual cost of auto insurance is 960 dollars. Assume the standard deviation is 194 dollars. You will use a simple random sample of 101 auto insurance policies.

Find the probability that a single randomly selected policy has a mean value between 927.2 and 1016 dollars.
P(927.2 < X < 1016) =

Find the probability that a random sample of size n=101n=101 has a mean value between 927.2 and 1016 dollars.
P(927.2 < M < 1016) =

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

P(927.2 < X < 1016 ) = P[(927.2 -960) / 194< (x - ) / < (1016 - 960) /194 )]

= P(-0.17 < Z <0.29 )

= P(Z <0.29 ) - P(Z < -0.17)

Using z table,  

= 0.6141 -0.4325

=0.1816
(B)

n = 101

m= 960

m =  / n = 194 / 101=19.3037

P(927.2 < M < 1016) = P[(927.2 -960) ) / 19.3037< (M - m ) / m < (1016 - 960) /19.3037 )]

= P( -1.70< Z < 2.90)

= P(Z < 2.90) - P(Z < -1.70)

Using z table,  

= 0.9981 -0.0446

=0.9535