Hi Salma,
This problem uses the classic equation in elementary statistics, z=(x-mu)/sigma. Let's break that down and then we'll apply to part (a)
z=z-score, explained below
x=x, value you were given in problem to compare
mu=population mean
sigma=population standard deviation
(a) Let's identify our variables for that equation in this problem:
x=976
mu=1029
sigma=219
z=(976-1029)/219
z= -0.24
Now, we can take that value to a z-table, found online or in any stats textbook, look at the column on the left for -0.2, look up top for 0.04. This gives the probability that z is less than -0.24. This is called standardizing, and it means:
P(Z< -0.24)=P(X<976)
From table:
P(Z< -0.24)=P(X<976)=0.4052
This is the probability that a single value falls below $976.
(b) Now, when we start taking samples, we need to change the equation a little bit. Instead of z=(x-mu)/sigma, it now becomes:
z=(x-mu)/SE where:
SE=standard error
Now, standard error of a sample mean has its own formula. It is:
SE=sigma/sqrt(n)
sigma=standard deviation
n=sample size
For our problem,
sigma=219
n=60
SE=219/sqrt(60)
SE=28.273
Now, we have all the variables we need:
x=976
mu=1029
SE=28.273
z=(976-1029)/28.273
z= -1.87
Return to the z-table. We will standardize again:
P(Z< -1.87) = P(X<976)
P(X<976) =0.0307
This is the probability that a 60-person sample will have a mean auto insurance cost less than 976 dollars. This is low, and it makes sense because it is unlikely that that many people will have such good deals on auto insurance. I hope this helps.
