1. Let x be a continuous random variable. What is the probability that x assumes a single value, such as a (use numerical value)?
2. The following are the three main characteristics of a normal distribution.
The total area under a normal curve equals _____.
A normal curve is ___________ about the mean. Consequently, 50% of the total area under a normal distribution curve lies on the left side of the mean, and 50% lies on the right side of the mean.
Fill in the blank. The tails of a normal distribution curve extend indefinitely in both directions without touching or crossing the horizontal axis. Although a normal curve never meets the ________ axis, beyond the points represented by µ  3σ to µ+ 3σ it becomes so close to this axis that the area under the curve beyond these points in both directions is very close to zero.
3. For the standard normal distribution, find the area within one standard deviation of the
mean that is, the area between μ − σ and μ + σ. Round to four decimal places.
4. Find the area under the standard normal curve. Round to four decimal places.
a) between z = 0 and z = 1.95
b) between z = 0 and z = −2.05
c) between z = 1.15 and z = 2.37
d) from z = −1.53 to z = −2.88
e) from z = −1.67 to z = 2.24
5. The probability distribution of the population data is called the (1) ________. Table 7.2 in the text provides an example of it. The probability distribution of a sample statistic is called its (2) _________. Table 7.5 in the text provides an example it.
Probability distribution
Population distribution
Normal distribution
Sampling distribution
6. ___________ is the difference between the value of the sample statistic and the value of the corresponding population parameter, assuming that the sample is random and no nonsampling error has been made. Example 71 in the text displays sampling error. Sampling error occurs only in sample surveys.
7. Consider the following population of 10 numbers. 20 25 13 19 9 15 11 7 17 30
a) Find the population mean. Round to two decimal places.
b) Rich selected one sample of nine numbers from this population. The sample included the numbers 20, 25, 13, 9, 15, 11, 7, 17, and 30. Calculate sampling error for this sample. Round to decimal places.
8. Fill in the blank. The F Distribution is ________ and skewed to the right. The F distribution has two numbers of degrees of freedom: df for the numerator and df for the denominator. The units of an F distribution, denoted by F, are nonnegative.
9. Find the critical value of F for the following. Round to two decimal places.
a) df = (3, 3) and area in the right tail = .05
b) df = (3, 10) and area in the right tail = .05
c) df = (3, 30) and area in the right tail = .05
10. The following ANOVA table, based on information obtained for three samples selected from three independent populations that are normally distributed with equal variances, has a few missing values.
Source of Variation 
Degrees of Freedom 
Sum of Squares 
Mean Square 
Value of the Test Statistic 
Between 
2 
II 
19.2813 

Within 
89.3677 
III 
F = ___V__ = VII VI 

Total 
12 
IV 
a) Find the missing values and complete the ANOVA table. Round to four decimal places.
b) Using α = .01, what is your conclusion for the test with the null hypothesis that the means of the three populations are all equal against the alternative hypothesis that the means of the three populations are not all equal?
Reject H_{0}. Conclude that the means of the three populations are equal.
Reject H_{0}. Conclude that the means of the three populations are not equal.
Do not reject H_{0}. Conclude that the means of the three populations are equal.
Do not reject H_{0}. Conclude that the means are of the three populations are not equal.
4. Find the area under the standard normal curve. Round to four decimal places.
a) between z = 0 and z = 1.95
P(0 < z < 1.95) = P(z < 1.95)  P(z < 0)
= 0.9744  0.5 = 0.4744
b) between z = 0 and z = −2.05
P(0 < z < 2.05) = P(z < 0)  P(z < 2.05)
= 0.5  0.0202
= 0.4798
c) between z = 1.15 and z = 2.37
P(1.15 < z < 2.37)
= P(z < 2.37)  P(z < 1.15)
= 0.1162
=NORM.DIST(2.37,0,1,TRUE)  NORM.DIST(1.15,0,1,TRUE)
d) from z = −1.53 to z = −2.88
P(2.88 < z < 1.53)
= P(1.53 < z)  P(2.88 < z)
= 0.0610
=NORM.DIST(1.53,0,1,TRUE)  NORM.DIST(2.88,0,1,TRUE)
e) from z = −1.67 to z = 2.24
P(1.67 < z < 2.24)
= P(2.24 < z)  P(1.67 < z)
= 0.9400
=NORM.DIST(2.24,0,1,TRUE)  NORM.DIST(1.67,0,1,TRUE)
Get Answers For Free
Most questions answered within 1 hours.