# Project #111716 - normal distribution

HOMEWORK 4

EXPECTED VALUE AND NORMAL DISTRIBUTIONS

Due: Saturday, February 27th

EXPECTED VALUES:                E(X) = å XiP(Xi)

PROBLEM 1 (8 points):

What is the expected value of a lottery ticket where there are only two chances in a million of winning the grand prize of \$25 Million?

PROBLEM 2 (8 points):

You have been offered a chance to purchase a lottery ticket with a 2% chance of making \$1,000; 5% chance of making \$100; and 93% chance of making \$0.  The price of the ticket is \$20.

PROBLEM 3 (9 points):

You have been offered a business deal.  You estimate that there is a 1% chance of making \$100,000; 4% chance of making \$40,000; a 20% chance of making \$10,000; and 75% chance of making \$0.  How much should you be willing to pay for this deal?  Do you think you would actually pay that much?  Why?

(SAMPLING) NORMAL DISTRIBUTIONS

General thoughts for this problem set:

Any normal distribution can be converted into a standard normal distribution by transforming the normal random variable into the standard normal random variable:

Z =

μ = the mean and σ = is the standard deviation.

For each question you need to ask yourself if you're looking for X or you’re looking for Z:

• You're looking for X if you're given a percentile and asked to come up with a concrete number (number of years or score on a test)
• You're looking for Z if you're given a concrete number (like a score or number of years) and asked to find out a probability or a percentile.  Based on the Z value from the table you can find the area under the curve.

Before solving the homework, again, carefully review the lecture notes.  It is important to also understand why you are doing the specific computations.  The next step would be for you to check and understand the solved problems from the lecture notes and the solution of the following problem.  It is always useful to draw the picture first.

I strongly suggest you to use the z-table of the Lecture Notes 4C.

PROBLEM :

Suppose that New York State high school average scores, for students who graduate, are normally distributed with a population mean of 70 and a population standard deviation of 13.

a) The “middle” 95% of all NYS high school students have average scores between ______ and ________?

b) What proportion of NYS high school students have average scores between 60 and 75?

c) Calculate the 14th percentile.

d)  Calculate the 92nd percentile.

SOLUTION

a) The “middle” 95% of all NYS high school students have average scores between ___ and ___ ?

.4750   .4750

?          70        ?          HS average

________________________________

-1.96      0          1.96     Z

If you want to consider the middle 95%, you will have to take 2.5% from the left tail and 2.5% from the right tail.

On each direction, from the middle (z = 0) there is a distance of 50% – 2.5% = 0.4750.  From the chart, to an area of .4750 corresponds a z =1.96

X = μ +(± Z)x σ = 70 + (± 1.96) x13 = 70 ± 1.96 (13)   è 70 ± 25.48

ANS: 95% of all NYS high school students have average scores between 44.52 and 95.48.

b) What proportion of NYS high school students has average scores between 60 and 75?

.2794      .1480

60        70        75        HS average

________________________________

-.77        0          .38       Z

Z1 = (60 – 70) / 13 = -0.77     (it corresponds to the area under the curve from the mean value to 60;                         the area on the left of the mean; it is the probability of having an average value between 60 and 70)

Z2 = (75 – 70) / 13 = .38        (it corresponds to the area under the curve from the mean value to 75;the area on the right of the mean; it is the probability of having an average score between 70 and 75)

We are interested on the total area    ANS: .2794 + .1480 = .4274

42.74% of NYS high school students have average scores between 60 and 75.

Percentiles:

Here is a definition of percentile from Wikipedia:
"In statistics, a percentile (or centile) is the value of a variable below which a certain percent of observations fall. For example, the 20th percentile is the value (or score) below which 20 percent of the observations may be found. "

A percentile represents a value.  If for example somebody's GPA score falls in the 93%ile  (93 percentile) it means that 93% of the students who took the test received a lower score.

It is important to acknowledge the fact that the mean represents the 50th percentile.  This will give you an idea in terms of where the other values for other specific percentiles fall.  For example any percentile > 50%, will be represented by a value greater than the mean and it will be (on the x axis) to the right of the mean value.  The normalized value (z-score) will be a positive value.

On the other hand any percentile < 50%, will be represented by a value smaller than the mean and it will be to the left of the mean value.  The normalized value (z-score) will be a negative value.

80th percentile
You need to find the X value corresponding to the 80%.
The 80 percentile is to the right of 50%ile.   The corresponding Z is positive, and X is greater than the mean.
How are we going to get this value?

Z = (X-miu)/stdev

In the above formula we know miu (mean), we know stdev, and we can compute Z.  Next step is the computation of X.
How do we find Z?  Z represents the distance from mean to our value X.
The distance from mean to X (representing 80%) will correspond to the area
under the curve  between 50% and 80% = .8 - .5= .3000

once we have the area we can retrieve Z and next we can compute X.

The reasoning is similar for let’s say the 27%; Now your X will be to the left of the mean and the corresponding Z should be considered negative.   X value will be lower than the mean.
The distance from mean to the X (representing 27%) will have as corresponding area  under the curve (the area between 50% and 27%) of  .5 - .27= .2300

c) Calculate the 14th percentile.

.1400   .3600

?          70                   HS average

________________________________

-1.08      0                     Z

We need to calculate X.  First we need to find the Z value.

In a normal distribution mean=median=mode.  The value 0 corresponds to the median = 50%.

This means that the 14% is on the left of 0 (the 50%).

What is the value of Z, the distance from 0?

The area under the curve will be 50% - 14% = 36% = 0.3600

0.3600 corresponds (from the table) to a Z of 1.08.  This Z is to the left of 0, therefore we will consider its negative value (-1.08)

Z = -1.08 = (X- 70) / 13

X =  55.96 years

d)  Calculate the 92nd percentile.

.4200

70        ?          HS average

________________________________

0          1.41     Z

Z=0 corresponds to the median = 50%.

This means that the 92% is on the right of 0 (the 50%).

The area under the curve, from 0 to Z is: 92%-50% = 42% = 0.4200  à (from the table) Z = 1.41

Z = 1.41 = (X- 70) / 13

X = 88.33 years

PROBLEM 4 (25 points):

Suppose CUNY professors have an average life, normally distributed, of 80 years with a population standard deviation of 9 years.

a)What percent of CUNY professors will live more than 86 years?

b) What percent of CUNY professors will not make it past the age of 50?

c) Calculate the 96th percentile.

d) Calculate the 5nd percentile.

e) What proportion of CUNY professors will live between 75 and 85 years?

PROBLEM 5 (20 points)

Suppose the lifetimes of Hoover vacuum cleaners are normally distributed with an average life (μ) of 12 years and a population standard deviation (σ) of 1.4 years.

a) What proportion of Hoover vacuum cleaners will last 14 years or more?

b) What proportion of Hoover vacuum cleaners will last 7 years or less?

c)  What proportion of Hoover vacuum cleaners will last between 11 and 13 years?

d)  Calculate the 85th percentile.

e)  Calculate the 7th percentile.

PROBLEM 6 (10 points)

Scores of high school seniors taking the English Regents examination in New York State are normally distributed with a mean of 70 and a standard deviation of 10.  Find the probability that a randomly selected high school senior will have a score between 70 and 80?

PROBLEM 7 (20 points)

The average individual monthly spending in US for paging and messaging services is \$10.50. If the standard deviation is \$2.5 and the amounts are normally distributed:

What is the probability that a randomly user pays more than \$9.00?

What is the probability that a user pays below \$10.00?

Compute the 75% percentile.

 Subject Mathematics Due By (Pacific Time) 02/27/2016 05:00 pm
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