Respuesta :
We have a population that is normally distributed with mean 991 and standard deviation 237.
If X is the random variable "annual cost of auto insurance", we can express it as X ~ N(991, 237).
Now, if we take a sample of size n = 40 of that population and define M as the "sample mean of annual cost of auto insurance", the distribution for M will be a normal ditribution with the same mean as X but with a standard deviation that will be affected by the sample size:
[tex]\sigma_M=\frac{\sigma}{\sqrt{n}}=\frac{237}{\sqrt{40}}\approx\frac{237}{6.32456}\approx37.4730[/tex]Then, the random variable M can be expressed as M ~ N(991, 37.4730).
We can now calculate the probability that one randomly selected auto insurance is less than $995.
To find that we calculate the z-score using the population distribution:
[tex]z=\frac{X-\mu}{\sigma}=\frac{995-991}{237}=\frac{4}{237}=0.01688[/tex]and then use it to calculate the probability as:
[tex]P(X<995)=P(z<0.01688)=0.5067[/tex]We now calculate the probability for the sample mean to be less than $995, with a sample size of n =40.
To do that we use the parameters of the sampling distribution:
[tex]z=\frac{M-\mu_M}{\sigma_M}=\frac{995-991}{37.4730}=\frac{4}{37.4730}=0.10674[/tex]and use it to calculate the probability as:
[tex]P(M<995)=P(z<0.10674)=0.5425[/tex]In this case, as the sample size is greater than 30, we can consider we have a simple big enough to conclude we don't need the assumption of normality for the underlying population distribution.
Answer:
a) N(991, 237)
b) N(991, 37.4730)
c) 0.5067
d) 0.5425
e) No, the assumption of normality is not necessary.
