Answer:
Point estimate = 76.4
Margin of Error = 2.680
Explanation:
Given that distribution is approximately normal;
The point estimate = sample mean, xbar = 76.4
The margin of error = Zcritical * s/√n
Tcritical at 95%, df = 42 - 1 = 41
Tcritical(0.05, 41) = 2.0195
Margin of Error = 2.0195 * (8.6/√42)
Margin of Error = 2.0195 * 1.327
Margin of Error = 2.67989
Margin of Error = 2.680