Question

CNNBC recently reported that the mean annual cost of auto insurance is 988 dollars. Assume the standard deviation is 105 dollars. You will use a simple random sample of 110 auto insurance policies.

Find the probability that a single randomly selected policy has a mean value between 994 and 1008 dollars.
P(994 < X < 1008) =

Find the probability that a random sample of size n=110n=110 has a mean value between 994 and 1008 dollars.
P(994 < M < 1008) =

Answer 1

Solution :

Given that ,

mean = = 988

standard deviation = = 105

P(994 < x < 1008) = P[(994 - 988)/ 105) < (x - ) /  < (1008 - 988) / 105) ]

= P(0.06 < z < 0.19)

= P(z < 0.19) - P(z < 0.06)

= 0.5753 - 0.5239

= 0.0514

P(994 < x < 1008) = 0.0514

= / n = 105 / 110 = 10.0114

= P[(994 - 988) /10.0114 < ( - ) / < (1008 - 988) / 10.0114)]

= P(0.60 < Z < 2.00)

= P(Z < 2.00) - P(Z < 0.60)

= 0.9772 - 0.7257

= 0.2515

P(994 < M < 1008) = 0.2515