7.43: Let
denote the random variable for height and
for the sample mean. Then if
is the mean of
So the probability that the difference between the sample and population means does not exceed 0.5 inch is

per the empirical or 68/95/99.7 rule.
7.44: For a sample of size <em>n</em>, the sample standard deviation would be
. We want to find <em>n</em> such that

Comparing to the equation from the previous part, this means we would need

so a sample of at least 157 men would be sufficient.
Answer: All four runners ran .96 seconds
Step-by-step explanation: The total time of the four runners, divided by the amount of runners.
3.86/4= .965
Its not a full questions...... send the full question
The first one is a maximum and the a.s. is x=1
The second one is a minimum and the a.s. is x=-2
2/3
divide by 17
34/17 is 2
51/17 is 3