Calculate the upper and lower limit for a 95% confidence interval about this mean.

a family needs a new car, but isn't sure they can fit the payment into their budget. a sample of 36 months of grocery bills yields a mean of $94 with a standard deviation of $10. if the upper limit of a 95% confidence level is below $100, the family can afford to buy the car.

standard error = (standard deviation)/(square root of sample size)

upper limit (dollars and cents)

lower limit (dollars and cents)

We have the mean, μ = 94 and standard deviation, σ = 10
We also have the number of samples, n = 36 and the confidence level, α = 95%

The formula for confidence interval is given

μ + z* (σ/√n) and μ - z* (σ/√n)

Where z* is the z-values for the confidence level

z* for  95% level of confidence is 1.96

Substitute this into the formula, we have

Upper limit ⇒ 94+(1.96)(10÷√36) = 94+(1.96)(10/6) = $97.27 
Lower limit ⇒ 94 - (1.96)(10÷√36) = 94 - (1.96)(10/6) = $90.73

I am pretty sure the answer is 24.65. Correct me if I am wrong. Hope this helps!