The mean annual cost of car insurance is $939. The standard deviation is $300. You have been asked to compute the probability that a simple random sample of car insurance policies of size 100 will have a sample mean within $25 of the population mean.
Compute the probability that a simple random sample of automobile insurance policies of size 100 will have a sample mean within $25 of the population mean.
a. .0638
b. .0319
c. .5934
d. .2967
e. .6920
show your calculations also please
Compute the probability that a simple random sample of automobile insurance policies of size 100 will have a sample mean within $25 of the population mean.
a. .0638
b. .0319
c. .5934
d. .2967
e. .6920
show your calculations also please
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939-25 = 914
939+25 = 964
Mean μ = 939
Standard deviation σ = 300
Standard error σ / √ n = 300 / √ 100 = 30
standardize xbar to z = (xbar - μ) / (σ / √ n )
P( 914 < xbar < 964) = P[( 914 - 939) / 30 < z < ( 964 - 939) / 30]
P( -0.8333 < z < 0.8333) = (0.2967)(2) = .5934
(from normal probability table)
939+25 = 964
Mean μ = 939
Standard deviation σ = 300
Standard error σ / √ n = 300 / √ 100 = 30
standardize xbar to z = (xbar - μ) / (σ / √ n )
P( 914 < xbar < 964) = P[( 914 - 939) / 30 < z < ( 964 - 939) / 30]
P( -0.8333 < z < 0.8333) = (0.2967)(2) = .5934
(from normal probability table)