1. For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right 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 skewed-right with mean = 10 minutes and standard error = 0.8 minutes. b) Distribution is skewed-right 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 skewed-right 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. Why is the Central Limit Theorem so important to the study of sampling distributions? a) It allows us to disregard the size of the sample selected when the population is not normal. b) It allows us to disregard the shape of the sampling distribution when the size of the population is large. c) It allows us to disregard the size of the population we are sampling from. d) It allows us to disregard the shape of the population when n is large. ANSWER: d 4. Suppose a sample of n = 50 items is drawn from a population of manufactured products...

...require that we know whether we have a sample or a population. 2. The following numbers represent the weights in pounds of six 7year old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45} Find the mean; median; mode; range; quartiles; variance; standard deviation. Solution: mean = 46.166.... median = 48 mode does not exist range = 35 Q1 = 45 Q2 = median = 48 Q3 = 51 variance = 112.1396 standard deviation =10.59 3. If the variance is 846, what is the standard deviation? Solution: standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data
34, 38, 22, 21, 29, 37, 40, 41, 22, 20, 49, 47, 20, 31, 34, 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6
This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot?
45 40 35 30 25 20 15 10 5 0 0 5 10 15 20
Solution: Weak positive linear correlation 6. What values can r take in linear regression? Select 4 values in this interval and describe how they would be interpreted. Solution: the values are between –1 and +1 inclusive. -1 means strong negative correlation +1 means strong positive correlation 0 means no correlation .5 means moderate positive correlation etc. 7. Does correlation imply causation? Solution: No.
8. What do we call the r value. Solution: The correlation coefficient....

...
NormalDistributionNormaldistribution is a statistics, which have been widely applied of all mathematical concepts, among large number of statisticians. Abraham de Moivre, an 18th century statistician and consultant to gamblers, noticed that as the number of events (N) increased, the distribution approached, forming a very smooth curve.
He insisted that a new discovery of a mathematical expression for this curve could lead to an easier way to find solutions to probabilities of, “60 or more heads out of 100 coin flips.” Along with this idea, Abraham de Moivre came up with a model that has a drawn curve through the midpoints on the top of each bar in a histogram of normally distributed data, which is called, “Normal Curve.”
One of the first applications of the normaldistribution was used in astronomical observations, where they found errors of measurement. In the seventeenth century, Galileo concluded the outcomes, with relation to the measurement of distances from the star. He proposed that small errors are more likely to occur than large errors, random errors are symmetric to the final errors, and his observations usually gather around the true values. Galileo’s theory of the errors were discovered to be the characteristics of normaldistribution and the formula for...

...under a Standard Normal curve
a) to the right of z is 0.3632;
b) to the left of z is 0.1131;
c) between 0 and z, with z > 0, is 0.4838;
d) between -z and z, with z > 0, is 0.9500.
Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table)
b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table)
c ) the area between 0 to z is 0.4838, z = 2.14
d) the area to the right of +z = ( 1-0.95)/2 = 0.025, therefore z = 1.96
3. Given the Normally distributed variable X with mean 18 and standard deviation 2.5, find
a) P(X < 15);
b) the value of k such that P(X < k) = 0.2236;
c) the value of k such that P(X > k) = 0.1814;
d) P( 17 < X < 21).
Ans : X ~ N ( 18, 2.52)
a) P ( X < 15)
P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places)
b) P ( X < k) = 0.2236
P ( Z < ( k – 18) / 2.5 ) = 0.2236
From normal table, 0.2236 = -0.76
(k-18)/2.5 = - 0.76, solve k = 16.1
c) P (X > k) = 0.1814
P ( Z > (k-18)/2.5 ) = 0.1814
From normal table, 0.1814 = 0.91
(k-18)/ 2.5 = 0.91, solve k = 20.275
d) P ( 17 < X < 21)
P ( (17 -18)/2.5 < Z < ( 21-18)/2.5)
P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places)...

...NORMALDISTRIBUTION
1. Find the
distribution:
a.
b.
c.
d.
e.
f.
following probabilities, the random variable Z has standard normal
P (0< Z < 1.43)
P (0.11 < Z < 1.98)
P (-0.39 < Z < 1.22)
P (Z < 0.92)
P (Z > -1.78)
P (Z < -2.08)
2. Determine the areas under the standard normal curve between –z and +z:
♦ z = 0.5
♦ z = 2.0
Find the two values of z in standard normaldistribution so that:
P(-z < Z < +z) = 0.84
3. At a university, the average height of 500 students of a course is 1.70 m; the standard
deviation is 0.05 m. Find the probability that the height of a randomly selected student is:
1. Below 1.75 m
2. Between 1.68 m and 1.78 m
3. Above 1.60 m
4. Below 1.65m
5. Above 1.8 m
4. Suppose that IQ index follows the normaldistribution with µ = 100 and the standard
deviation σ = 16. Miss. Chi has the IQ index of 120. Find the percentage of people who
have the IQ index below that of Miss. Chi.
5. The length of steel beams made by the Smokers City Steel Company is normally
distributed with µ = 25.1 feet and σ = 0.25 feet.
a. What is the probability that a steel beam will be less than 24.8 feet long?
b. What is the probability that a steel beam will be more than 25.25 feet
long?
c. What is the probability that a steel beam will be between 24.9 and 25.7
feet long?
d. What is the probability that a steel beam will be between 24.6 and 24.9
feet long?
e....

...6
The NormalDistribution
Objectives
Outline
After completing this chapter, you should be able to
1
2
3
Identify distributions as symmetric or skewed.
4
Find probabilities for a normally distributed
variable by transforming it into a standard
normal variable.
Introduction
6–1
NormalDistributions
Identify the properties of a normaldistribution.
Find the area under the standard normaldistribution, given various z values.
5
Find speciﬁc data values for given
percentages, using the standard normaldistribution.
6
6–3 The Central Limit Theorem
6–4 The Normal Approximation to the Binomial
Distribution
Use the central limit theorem to solve
problems involving sample means for large
samples.
7
6–2 Applications of the NormalDistribution
Use the normal approximation to compute
probabilities for a binomial variable.
Summary
6–1
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Confirming Pages
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Chapter 6 The NormalDistribution
Statistics
Today
What Is Normal?
Medical researchers have determined so-called normal intervals for a person’s blood
pressure, cholesterol, triglycerides, and the...

...Math 221
Week 6 Lab
Submitted by: Merima Ceric
Part 1. NormalDistributions and Birth Weights in America
1) What percent of the babies born with each gestation period have a low birth weight (under 5.5 pounds)?
a) Under 28 = 99.88%
The NORMDIST formula was used to calculate: =NORMDIST(5.5,1.88,1.99,True)
X= 5.5
Mean= 1.88
Standard Deviation=1.19
b) 32 to 35 weeks = 43.83%
The NORMDIST formula was used to calculate: =NORMDIST (5.5,5.73,1.48,True)
X= 5.5
Mean= 5.73
Standard Deviation=1.48
c) 37 to 39 weeks = 4.66%
The NORMDIST formula was used to calculate: =NORMDIST(5.5,7.33,1.09, True)
X= 5.5
Mean= 7.33
Standard Deviation=1.09
d) 42 weeks and over = 2.75%
The NORMDIST formula was used to calculate:=NORMDIST( 5.5, 7.65,1.12, True)
X= 5.5
Mean= 7.65
Standard Deviation=1.12
2) Describe the weights of the top 10% of the babies born with each gestation period.
a) 37 to 39 weeks = 8.73 lbs
The NORMINV formula was used to calculate: =NORMINV(0.9,7.33,1.09)
Probability= 0.9
Mean=7.33
Standard Deviation= 1.09
b) 42 weeks and over= 9.09 lbs
The NORMINV formula was used to calculate: =NORMINV(0.9,7.65,1.12)
Probability= 0.9
Mean=7.65
Standard Deviation= 1.12
3) For each gestation period, what is the probability that a baby will weigh...

... 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 non-normal). 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....

...NormalDistribution:- A continuous random variable X is a normaldistribution 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.
Normaldistribution 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
NormalDistribution 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...