Download 1. The total area under the curve of the standard normal distribution

1. The total area under the curve of the standard normal distribution is not necessarily 1.0. True statement?
Yes. The statement is True.
[Since the standard normal curve crosses the horizontal axis only at -infinity and +infinity, the total area under the curve
approaches 1.0, but for all practical purposes, it is taken as 1.0.]
2.The area to the right of z = 1.52 is 0.4357
A) True
B) False
The statement is False, since the area is 0.0642
3.In which of the following binomial distributions is the normal approximation appropriate?
A) n = 50, p = 0.01
B) n = 500, p = 0.001
C) n = 100, p = 0.05
D) n = 50, p = 0.02
Of the given options, in (C) we have p = 0.05, which is not too close to 0. The sample size is also quite large (100).
Therefore, in this case, using the normal distribution will be appropriate.
4.A normal distributed population has a mean of 250 pounds and a standard deviation of 10 pounds. Given n = 20,
what is the probability that this sample will have a mean value between 245 and 255 pounds?
A) 0.9750
B) 0.4875
C) 0.3830
D) 0.0876
z1 = (x1 - m)/(s/sqrt n) = (245 - 250)/(10/sqrt 20) = -2.2361
z2 = (x2 - m)/(s/sqrt n) = (255 - 250)/(10/sqrt 20) = 2.2361
P(x1 < x < x2) = P(z1 < z < z2) = 0.9746
Suppose the annual consumption of chicken mean is 20.84 pounds per person, and that the standard deviation for the
consumption of chicken per person is 9.193 pounds. The mean weight of chicken consumed for a sample of 200
randomly selected people is one value of many that form the sampling distribution of sample means.
1) Describe the shape of this sampling distribution. 2) What is the mean value for this sampling distribution? 3) What
is the standard deviation of this sampling distribution?
(1) Since the sample size is 200 (> 30), the shape of the sampling distribution is bell-shaped. It is normal distribution.
(2) Since the distribution is normal, the mean of the sampling distribution = the population mean = 20.84 pounds
(3) Standard deviation of the sampling distribution = Population standard deviation/sample size
= 9.193/200 = 0.65.