A population of measurements is approximately normally distributed with mean of 25 and a variance of 9. Find the probability that a measurement selected at random will be between 19 and 31. Solution: The values 19 and 31 must be transformed into the corresponding z values and then the area between the two z values found. Using the transformation formula from X to z (where µ = 25 and σ √9 = 3), we have
z19 = (19 – 25) / 3 = 2 and z31 = (31  25) / 3 = +2
From the area between z =±2 is 2(0.4772) = 0.9554
Therefore the probability that a measurement selected at random will be between 19 and 31 is about 0.95. This area (probability) is shown fir the X values and for the z values.
σ = 3 0.95σ = 1 0.95
X
19 25 312 0 +2
Normal curve showing Standard normal curve showing
area between 19 and 31area between 2 and +2
Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100.
Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this university?
Solution:
Let x be the random variable that represents the scores. x is normally distributed with a mean of 500 and a standard deviation of 100. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve by 100, we obtain percentages.
For x = 585, z = (585  500) / 100 = 0.85
The proportion P of students who scored below 585 is given by
P = [area to the left of z = 0.85] = 0.8023 = 80.23%
Tom scored better than 80.23% of the students who took the test and he will be admitted to this University.
... standard deviation, variance, standard error of the mean, and confidence intervals. These statistics are used to summarize data and provide information about the sample from which the data were drawn and the accuracy with which the sample represents the population of interest. The mean, median, and mode are measurements of the “central tendency” of the data. The range, standard deviation, variance, standard error of the mean, and confidence intervals provide information about the “dispersion” or variability of the data about the measurements of central tendency.
MEASUREMENTS OF CENTRAL TENDENCY The appropriateness of using the mean, median, or mode in data analysis is dependent upon the nature of the data set and its distribution (normal vs nonnormal). The mean (denoted by x) is calculated by dividing the sum of the individual data points (where Σ equals “sum of”) by the number of observations (denoted by n). It is the arithmetic average of the observations and is used to describe the center of a data set.
mean=x= One of the most basic purposes of statistics is simply to enable us to make sense of large numbers. For example, if you want to know how the students in your school are doing in the statewide achievement test, and somebody gives you a list of all 600 of their scores, that’s useless. This everyday problem is even more obvious and staggering when you’re dealing, let’s say, with the population data for the nation....
...time is known to have a skewedright distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights. a) Distribution is skewedright with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewedright with mean = 10 minutes and standard error = 8 minutes. c) Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes. d) Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes. ANSWER: c 2. Suppose the ages of students in Statistics 101 follow a skewedright distribution with a mean of 23 years and a standard deviation of 3 years. If we randomly sampled 100 students, which of the following statements about the sampling distribution of the sample mean age is incorrect? a) The mean of the sampling distribution is equal to 23 years. b) The standard deviation of the sampling distribution is equal to 3 years. c) The shape of the sampling distribution is approximately normal. d) The standard error of the sampling distribution is equal to 0.3 years. ANSWER: b 3....
...SAMPLING DISTRIBUTIONS
6.1 POPULATION AND SAMPLING DISTRIBUTION 
6.1.1 Population Distribution 
Suppose there are only five students in an advanced statistics class and the midterm scores of these five students are:
70 78 80 80 95
Let x denote the score of a student.
• Mean for Population
Based on Example 1, to calculate mean for population:
[pic]
• Standard Deviation for Population
Based on example 1, to calculate standard deviation for population:
[pic]
6.1.2 Sampling Distribution 
▪ Sample statistic such as median, mode, mean and standard deviation
6.1.2.1 The Sampling Distribution of the Sample Mean
Reconsider the population of midterm scores of five students given in example 1. Let say we draw all possible samples of three numbers each and compute the mean.
Total number of samples = 5C3 =[pic]
Suppose we assign the letters A, B, C, D and E to scores of the five students, so that...
...6 More Inference about a Population
1. Inference about a Normal Population Mean: Variance Unknown 2. Inference about a Normal Population Variance 3. Inference about a Population Proportion Based on Large Samples Section 6.1 Inference about a Normal Population Mean: Variance Unknown Assumption The population follows 2 N(, ). Let X be the sample mean of a sample taken from the population, S be the sample standard deviation and n be the sample size. It can be X follows the t distribution proved that S/ n with n 1 degrees of freedom.
BS06 1
This graph shows 0.4 the probability N(0, 1) density functions of 0.3 N(0, 1) and some t(20) t(n), the t distribution with n 0.2 degrees of freedom where n is a 0.1 t(5) positive integer. t(1)
3 2 1 0 1 2 3
Interval estimation The 100(1 )% confidence interval for the population mean is s s , x t / 2, n1 x t / 2, n1 n n where the value t, n can be obtained from the t distribution table. Comparing with the case with known , here we use s instead of and t2,n1 instead of z2.
BS06 2
Density function of the t distribution with n degrees of freedom Area
t,n Example 1 A paint manufacturer wants to determine the average drying time of a new brand of interior wall paint. If for 12 test areas of equal size he obtained a mean drying time of 66.3 minutes and a standard deviation of 8.4...
...
5. A score that is likely to fall into the middle 68% of scores of a normaldistribution will fall inside these values: (Points : 1) 
. +/ 3 standard deviations
+/ 2 standard deviations
+/ 1 standard deviation
semiquartile range

6. It is important to assess the magnitude or strength of a relationship because this assists you with deciding whether or not a variable A causes variable B. (Points : 1) 
True
False

7. In a negative relationship, as the score of one variable decreases, the score on the second variable decreases. (Points : 1) 
True
False

8. A set of subjects, usually randomly sampled, selected to participate in a research study is called: (Points : 1) 
Population
Sample
Mode Group
Partial Selection

9. A perfect negative relationship between two variables is expressed as r=0. (Points : 1) 
True
False

10. When examining the relationship between a nominal variable and an interval or ratio variable, you would create a table using the nominal variables, calculate the mode and median of the interval or ratio variable, then make a decision regarding the relationship using the mode and median. (Points : 1) 
True
False

11. A Z score of +/1.96 is equivalent to these values on a normaldistribution. (Points : 1)...
...ORMAL(0, t) distribution.
A Wiener process with initial value W0 = x is gotten by adding x to a standard Wiener
process. As is customary in the land of Markov processes, the initial value x is indicated
(when appropriate) by putting a superscript x on the probability and expectation operators. The term independent increments means that for every choice of nonnegative real
numbers 0 ≤ s1 < t1 ≤ s2 < t2 ≤ · · · ≤ sn < tn < ∞, the increment random variables
Wt1 − Ws1 , Wt2 − Ws2 , . . . , Wtn − Wsn
are jointly independent; the term stationary increments means that for any 0 < s, t < ∞
the distribution of the increment Wt+s − Ws has the same distribution as Wt − W0 = Wt .
In general, a stochastic process with stationary, independent increments is called a L´vy
e
process; more on these later. The Wiener process is the intersection of the class of Gaussian
processes with the L´vy processes.
e
It should not be obvious that properties (1)–(4) in the deﬁnition of a standard Brownian
motion are mutually consistent, so it is not a priori clear that a standard Brownian motion
exists. (The main issue is to show that properties (3)–(4) do not preclude the possibility
of continuous paths.) That it does exist was ﬁrst proved by N. W IENER in about 1920.
´
His proof was simpliﬁed by P. L E VY; we shall outline L´ vy’s construction in section ??
e
below. But notice that properties (3) and (4) are compatible. This follows from...
...STATISTICS FOR MANAGEMENT
B1731
4
60
Note: Answer all questions. Kindly note that answers for 10 marks questions should be
approximately of 400 words. Each question is followed by evaluation scheme.
Questions
Marks
Total Marks
Q.No
1
Distinguish between Classification and Tabulation. Explain the structure and
components of a Table with an example.
Meaning of Classification and Tabulation
Differences between Classification and Tabulation
2
Structure and Components of a Table with an example
2
2
10
6
a) Describe the characteristics of Normal probability distribution.
b) In a sample of 120 workers in a factory, the mean and standard deviation of wages
were Rs. 11.35 and Rs.3.03 respectively. Find the percentage of workers getting wages
between Rs.9 and Rs.17 in the whole factory assuming that the wages are normally
distributed.
Characteristics of Normal probability distribution
Formula/Computation/Solution to the problem
3
4
10
6
a) The procedure of testing hypothesis requires a researcher to adopt several steps.
Describe in brief all such steps.
b) Distinguish between:
i. Stratified random sampling and Systematic sampling
ii. Judgement sampling and Convenience sampling
Hypothesis testing procedure
4
Differences
10
6
(3 each)
4
a) What is regression analysis? How does it differ from correlation analysis?
b) Calculate Karl Pearson’s...
...the probability that they have a mean pregnancy between 266 days and 268 days. A) 0.5517 B) 0.3577 C) 0.2881 D) 0.7881 Answer: B
13)
2
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 14) Find the zscore that corresponds to the given area under the standard normal curve. 14)
Answer: z = 0.42 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 15) If one card is drawn from a standard deck of 52 playing cards, what is the probability of drawing a red card? 1 1 1 1 A) B) C) D) 4 2 52 13 Answer: B 16) Find the standardized test statistic t for a sample with n = 10, x = 13.3, s = 1.3, and H0 : µ 14.2. Round your answer to three decimal places. A) 2.617 B) 3.010 C) 3.186 Answer: D 17) Find the area of the indicated region under the standard normal curve. 17)
= 0.05 if
15)
16)
D) 2.189
A) 0.0968 Answer: C
B) 0.0823
C) 0.9032
D) 0.9177
18) A sample of candies have weights that vary from 2.35 grams to 4.75 grams. Use this information to find the upper and lower limits of the first class if you wish to construct a frequency distribution with 12 classes. A) 2.352.65 Answer: B 19) Assume that the heights of men are normally distributed. A random sample of 16 men have a mean height of 67.5 inches and a standard deviation of 1.4 inches. Construct a 99% confidence interval for the population standard...