Question 1162123
a. the sum is 68 and 68/30=2.27 hours

b. the sd is 1.5/sqrt(n) for a sample of n teenagers.  The sample mean's point estimate would be 4 hours. I am assuming this is not relating to the above data.  If it is, do a 1-var stats on the data

c. P(1)=e^(-5)*5^1/1!=5e^-5)=0.0337

d.  p hat is 0.2 (50/250).  The half-interval is z*sqrt (p*(1-p)/250)=sqrt (0.2*0.8/250)=1.96*0.0253=0.0496 rounding at end.  The 95% CI is (0.15, 0.25) proportion of units defective.