Yearly healthcare expenses for a family of four are normally distributed with a mean expense equal to $3,000 and a standard deviation equal to $500. A sample of 36 families was selected and the mean and standard deviation were was found to be $3250 and $400 respectively. What is the probability of healthcare expenses in the population being greater than $4,000?

We are given that yearly healthcare expenses for a family of four are normally distributed with a mean expense equal to $3,000 and a standard deviation equal to $500.

Let X = yearly healthcare expenses of a family

The z-score probability distribution for the normal distribution is given by;

Z = [tex]\frac{ X-\mu}{\sigma} }[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population mean expense = $3,000

[tex]\sigma[/tex] = standard deviation = $500

Now, the probability of healthcare expenses in the population being greater than $4,000 is given by = P(X > $4,000)

P(X > $4,000) = P( [tex]\frac{ X-\mu}{\sigma} }[/tex] > [tex]\frac{4,000-3,000}{{500}{ } }[/tex] ) = P(Z > 2) = 1 - P(Z [tex]\leq[/tex] 2)

= 1 - 0.97725 = 0.02275

The above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.97725.