D. z(70000) = (70000-60000)/2000 = 5
prob(z > 5)
= about 0% (it's greater than 0, but the z table shows it as 0)
It is not reasonable, since the proportion of a population that's more than 5 standard deviations above the mean is basically 0.
Chap. 8
Ex. 38
A. z = (25-23.5)/(5/sqrt(50))
z = 2.12132
prob(z > 2.12132)
= 0.0169
B. z(22.5) = (22.5-23.5)/(5/sqrt(50)) = -1.4142
prob(-1.4142 < z < 2.12132)
= 0.9044
C. z = +/-1.6449
The interval goes from:
mean - z*sd/sqrt(N) to mean + z*sd/sqrt(N)
23.5 - 1.6449*5/sqrt(50) to 23.5 + 1.6449*5/sqrt(50)
22.3369 to 24.6631
Chap. 9
Ex. 54
A. The t value for df = n-1 = 19, with 90% confidence is:
1.7291
The interval goes from:
mean - t*sd/sqrt(N) to mean + t*sd/sqrt(N)
10979 - 1.7291*1000/sqrt(20) to 10979 - 1.7291*1000/sqrt(20)
B. The z value for 99% confidence is 2.5758
The formula for sample size is:
N = (z*sd/E)^2
N = (2.5758*1000/250)^2
N = 106.16
Round up to:
N = 107
Chap. 10
Ex. 42
H0: game length is >= 3.5 hours
Ha: game length is < 3.5 hours
mean = 2.9553
stdev = 0.5596
Get the t test statistic:
t = (x-mu)/(stdev/sqrt(N))
t = (2.9553-3.5)/(0.5596/sqrt(17))
t =...