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.
...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...
...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...
...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:...
...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...
...probably not attributable to chance is: (Points : 1) 
Type I error
Type II error
Statistical significance
In the semiquartile range

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...
...real numbers t with the following
properties:
(1)
(2)
(3)
(4)
W0 = 0.
With probability 1, the function t → Wt is continuous in t.
The process {Wt }t≥0 has stationary, independent increments.
The increment Wt+s − Ws has the N 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...
...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...
...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...