Normal Distribution: A continuous random variable X is a normal distribution with the parameters mean and variance then the probability function can be written as f(x) =  < x < ,  < μ < , σ > 0.
When σ2 = 1, μ = 0 is called as standard normal.
Normal distribution problems and solutions – Formulas:
X < μ = 0.5 – Z
X > μ = 0.5 + Z
X = μ = 0.5
where,
μ = mean
σ = standard deviation
X = normal random variable
Normal Distribution Problems and Solutions – Example Problems: Example 1:
If X is a normal random variable with mean and standard deviation calculate the probability of P(X<50). When mean μ = 41 and standard deviation = 6.5 Solution:
Given
Mean μ = 41
Standard deviation σ = 6.5
Using the formula
Z =
Given value for X = 50
Z =
=
= 1.38
Z = 1.38
Using the Z table, we determine the Z value = 1.38
Z = 1.38 = 0.4162
If X is greater than μ then we use this formula
X > μ = 0.5 + Z
50 > 41 = 0.5 + 0.4162
P(X) = 0.5 + 0.4162
= 0.9162
Example 2:
If X is a normal random variable with mean and standard deviation calculate the probability of P(X< 37). When mean μ = 20 and standard deviation = 15 Solution:
Given
Mean μ = 20
Standard deviation σ = 15
Using the formula
Z =
Given value for X = 37
Z =
=
= 1.13
Z = 1.13
Using the Z table, we determine the Z value = 1.13
Z = 1.13 = 0.4332
If X is greater than μ then we use this formula
X > μ = 0.5 + Z **** 37 > 20 = 0.5 + 0.3708 **** P(X) = 0.5 + 0.3708 = 0.8708
...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...
...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...
...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...
...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...