The Normal Distribution
Notes and review problems
The Normal Distribution
• A normal variable is continuous, ranges
from – to +, and forms a symmetrical
distribution, with a mean m and a standard
deviation s
m
The Normal Distribution
• The density function f(x) is
f (x)
1
e
2πs
1  x μ 
 

2 σ 
2
f(x)
f(x)
m
Solve the following problems
1.
Calculate the following
probabilities:
•
•
•
•
•
•
•
2.
P(0<Z<1.5)
P(Z<1.5)
P(-1.4<Z<.6)
P(Z>2.03)
P(Z>-1.44)
P(1.14<Z<2.43)
P(-.91<Z<-.31)
3.
Find Z0
•
•
•
for which 4% of the population
is located above it.
for which 2% of the population
is located below it.
that satisfies the condition:
P(-Z0<Z<+Z0) = .65
X is normally distributed with mean 100 and standard
deviation 20. What is the probability that X is:
1.
greater than 145?
2.
3.
4.
Less than 120
Less than 95
Between 75 and 100
Solutions
Set 1:
•
•
•
P(0<Z<1.5) = P(-<Z<1.5) – 0.5 = .9332 – 0.5 = 0.4332
With Excel use: =normsdist(1.5) – 0.5.
P(Z < 1.5) = P(-<Z<1.5) = 0.9332
P(-1.4 < Z < 0.6) = P(Z < 0.6) – P(Z < -1.4). Several steps are needed if using the Z
table from the text:
–
–
–
•
•
•
•
P(Z < 0.6) = 0.7257
P(Z < -1.4) = P(Z > +1.4) = 1 – P(Z < 1.4) = 1 – 0.9192 = .0808
P(-1.4 < Z < 0.6) = 0.7257 – 0.0808 = .6449
With Excel use: =normsdist(0.6) – normsdist(-1.4)
P(Z > 2.03) = 1 – P(Z < 2.03) = 1 – 0.9788 = .0212
With Excel use: = 1 – normsdist(2.03)
P(Z > -1.44) = P(Z < 1.44) = 0.9251
P(1.14 < Z < 2.43) = P(Z < 2.43) – P(Z < 1.14) = 0.9925 – 0.8508 = .1417
P(-.91 < Z < -.31) = P(Z < -.31) – P(Z < -.91) = P(Z < +.91) – P(Z < +.31) = 0.8186 –
0.6217 = .1969
With Excel use: =nornsdist(-.31) – normsdist(-.91).
Solutions
Set 2:
•
•
•
P(Z > Z0) = .04. Thus, P(Z < Z0) = 0.96. The corresponding Z0= 1.75
With Excel use: =normsinv(.96)
P(Z < Z0) = .02. The value of Z0 must be negative, therefore, if we want to
use the text table we need to use symmetry. Instead of finding Z0 as
requested, we’ll first find –Z0, which of course has a positive value! For
convenience let’s call this value Z1
P(Z > Z1) = .02. So, P(Z < Z1) = 0.98. Thus Z1 = 2.055 (or so). This means
that the value we are seeking Z0 = -2.055.
P(-Z0 < Z < Z0) = 0.65. Then P(Z < Z0) – P(Z <- Z0) = 0.65. However since
P(Z < -Z0) = P(Z > Z0) = 1 – P(Z < Z0), we have:
P(-Z0 < Z < Z0) = P(Z < Z0) – [1 – P(Z < Z0)] = 2P(Z < Z0) – 1 = 0.65. From
here we have P(Z < Z0) = 0.825, which yields Z0 =~0.935.
Another solution: Note that P(Z<-Z0)+ P(-Z0 < Z < Z0) +P(Z>Z0)=1, but
P(Z<-Z0)=P(Z>Z0) so, P(Z<-Z0)=.175, and –Z0 can be found.
Solutions
Set 3:
•
•
•
•
P(X>145) = P(Z>(145 – 100)/20) = P(Z>2.25). Use the Z-table.
With Excel use: =1 – normdist(145,100,20,True)
P(X<120) = P(Z<(120 – 100)/20) = P(Z<1). Use the table.
With Excel use: =normdist(120,100,20,True)
P(X<95) = use the same approach as before
P(75<X<100) = P((75 – 100)/20<Z<(100 – 100)/20). Use the table.
With Excel use: =normdist(100,100,20,True) – normdist(75,100,20,True)
Solve the following problems
4. The lifetime of light-bulbs that are advertised to last for 5000 hours
are normally distributed with a mean of 5,100 hours and a standard
deviation of 200 hours.
 What is the probability that a bulb lasts longer than the advertised
figure?
 If we wanted to be sure 98% of all the bulbs last longer than an
advertised figure, what figure should be advertised?
5. Because of high interest rates, most consumers attempt to pay-off
the credit card bills promptly. However this is not always possible. If
the amount of interest paid monthly by card-holders is normally
distributed with a mean of $27 and a standard deviation of $7,
• What proportion of the cardholders pay more than $30 in interest?
• What proportion of the cardholders pay less than $15 in interest?
• What interest payment is exceeded by only 20% of the cardholders?
Solutions
4.
Let X represent the bulb lifetime.
•
•
P(X>5100) = P(Z>(5000 – 5100)/200) = P(Z>-0.5) = P(Z<0.5)
Let the advertised figure be X0.
P(X>X0) = .98; X0 must reside to the left of m, therefore after transforming X
into a Z variable, the latter must have a negative value. Let us call this Z-value
–Z0 where –Z0 = (X0 – m)/s. Thus P(Z> –Z0) = .98 (click to see the graph), so
P(Z<-Z0) = .02. From the table we have –Z0 = -2.055 (the mid-value between -2.05 whose probability is 0.0202, and 2.06 whose probability is .0197). So,
the -Z0 = -2.055. From here we determine X0 by the equation -2.055 = (X0 –
5100)/200, or X0=5100 – 2.055(200) = 4689. The advertised lifetime should be
“4689 hours” if it is desired that 98% of the bulbs will last at least that long.
0.98
-Z0
Solutions
5.
Let X be the amount of interest paid monthly.
•
•
•
P(X>30) = P(Z>(30-27)/7) = P(Z>.42857). With Excel we have:
= 1-normdist(30,27,7,true) =.334118
P(X<15) = P(Z<(15 –27)/7) = P(Z<-1.71429) = P(Z>1.71429) =
1-P(Z<1.71429)=1-normsdist(1.71429)=0.043238
Let X0 be the interest exceeded by 20% of the cardholders.
P(X>X0) = 0.2; P(Z>Z0) = 0.2, where Z0 = (X0 – m)/s. Since Z0 is positive, to
use the table we look at P(Z<Z0)=.80. The corresponding value of Z0 is 0.84.
With Excel we use the function norminv or normsinv. These functions provide
the X0 or the Z0 respectively for a left hand tail probability. Observe:
X0: =norminv(.80,27,7) = 32.89; Z0 =normsinv(.80) = .8416 (to find X0 from Z0
note that X0 = m+Z0s = 27 + .8416(7) = 32.89).
0.20
Z0
Sampling Distributions
Notes and review problems
Sampling Distribution of the
Sample Mean
• The sample mean is normally distributed if the parent
distribution is normal.
• Then sample mean is approximately normally
distributed if the sample is sufficiently large (n30)
even if the parent distribution is not normal. The larger
the sample the better the approximation.
• The mean of the sample mean is the same as the
parent population mean (μx  μx )
σx
σ

• The standard deviation of the sample mean is x
n
Solve the following problems
1.
A sample of n=16 observations is drawn from a normal population with
m=1,000 and s=200. Find the following probabilities:
P( x  1,050)
P( x  960)
P( x  1,100)
2.
The heights of North American women are normally distributed with a mean of
64 inches and a standard deviation of 2 inches.
•
•
•
3.
What is the probability that a randomly selected woman is taller than 66 inches?
A random sample of 40 women is selected. What is the probability that the sample
mean height is greater than 66 inches?
If the population of heights is not normally distributed, would you change your answer
to part 2?
The manufacturer of cans of salmon claims the average can weight is 6.05
ounces. He further states the actual weight per can is a normal variable with a
mean of 6.05 and standard deviation of 0.18 ounces. A random sample of 36
observations is drawn.
a.
b.
What is the probability the sample mean weight is less 5.97?
If the random sample indeed produced a mean weight less than 5.97. Comment on
the statement made by the manufacturer.
Solutions
•
Problem 1
P( x  1,050)  P(Z  (1050 - 1000)/(200/ 16 ))  P(Z  1)
P( x  960)  P(Z  (960 - 1000)/(200/ 16 ))  P(Z  0.8)
•
P( x  1,100)  P(Z  (1100 - 1000)/(200/ 16)  P(Z  2)
Problem 2
Let X be a woman height.
P(x  66)  P(Z  (66 - 64)/2)  P(Z  1)
P( x  66)  P(Z  (66 - 64)/(2/ 40) )
•
Problem 3
P( X  5.97)  P(Z  (5.97  6.05)/(0.18/ 36 ))  P(Z  2.667)  1 P(Z  2.667)  .00383
The manufacturer is probably wrong. The chance to have a sample mean as
small as 5.97 is extremely small if the mean is 6.05. So if we actually have this
sample mean, the real population mean must be smaller.
Solve the following problems
4.
A portfolio can be composed of either one of two stocks, or of both. The mean and
standard deviation of the rate of return for each stock are .12 and .05. The returns
are normally distributed
a.
b.
If an investor decides to invest in one stock only, what is the probability that the
portfolio loses money? Earns more than 20%?
if the investor decides to invest in both stocks, one dollar in each stock, and assuming
the return on each stock is independent on the other
•
•
•
c.
d.
what is the probability that the portfolio loses money?
what is the probability that the portfolio earns more than 20%?
what do the above results tell you about diversification (both advantage and disadvantage)?
Now assume the return and standard deviation for stock 2 are .24 and .15 respectively.
Answer again parts ‘a’ and ‘b’. For part ‘b’ the portfolio return and standard deviation
depend on the proportion of the portfolio invested in each stock. Assume for simplicity
that the ratio is as before - a dollar for dollar.
Try to resolve part ‘c’ when the ratio of investments in the two stocks for the portfolio is
2 dollars in stock 1 for 1 dollar in stock 2. Note that the return in this case is
.67R1+.33R2 for each dollars invested in the portfolio. From this relationship you can
determine the mean and standard deviation of the portfolio.
Solutions
4.
Let X be the rate of return for each stock
a.
P(X<0) = P(Z<(0 - .12)/.05) = P(Z<-2.4) =
P(X>.20) = P(Z>(.20 - .12)/.05) = P(Z>1.6) =
Let X1 and X2 be the return for each stock.
b.
•
•
c.
d.
The portfolio loses money when X1 + X2 < 0. To calculate the probability
P(X1 + X2<0) we use the sampling distribution of the sample mean. We can do it
because both investments have the same mean and standard deviation, so we can
treat them as a sample of size n = 2:
P(X1+X2<0) = P((X1+X2)/2<0)=P(Z<(0 – .12)/(.05/21/2) =
P(Z<-3.3936) = close to zero.
P(X1+X2>.20) = P(Z>(.20-.12)/(.05/21/2) = P(Z>2.262)
We’ll not repeat part a, being very similar to the one solved before. For part b
we continue as follows: X = X1 + X2, so mx = m1+m2 =.12+.24 =.36, and sx2 =
s12 +s22=.052+.152 =.025. So the standard deviation for X is sx = .0251/2 =
.1581. From this we have:
P(X<0) = P(Z<(0 - .36)/(.1581) = P(Z<-2.27)
Repeat part C, but this time
mx =.67(.12)+.33(.24), and sx2 = (.672)(.052)+(.332)(.152)
Solve the problems
The signs on an elevator states “Maximum Capacity 1,140
Kilogram” or 16 persons”. If the mean weight is 75 kilos and the
standard deviation of the weight is 10 kilos,
5.
•
•
What is the probability that a sample of 16 persons would weigh more
than 1140 kilos.
What is the probability that 18 individuals will have a weight of more
than 80 Kilos out of this sample of 36 individuals?
Solutions
5.
a. P(X1+X2+…+X16>1140)= P X  1140   P Z  71.75  75   P(Z  1.5)
16 
10 / 16 


b. We use the normal distribution to find P(X>80). Then we use the
binomial distribution to calculate P(Binomial = 18), where a success
is defined as “the individual weight is more than 80 pounds”.
P(X>80) = P(Z>(80 – 75)/10) = P(Z>.5) = .3085. This is the
probability of a success for the binomial distribution we are about to
use. Define B as a binomial variable with n = 36 and p = .3085.
P(B=18) = binomdist(18,36,.3085,false)= 0.00762.
If you are unfamiliar with the function “binomdist” in Excel
you can use the normal approximation to the binomial (since you’ll
probably not find a binomial table for n=36). The mean of this normal
distribution is m =np=36(.3085)=11.12, and the standard deviation is
s =[np(1-p)]1/2=[(36)(.3085)(1-.3085)]1/2=2.77. The probability to
calculate is P(B=18)@P(17.5X18.5), where X is the normal
variable with the pre-calculated m and s. The result is .006715
Sampling Distribution of the
Sample Proportion (large samples)
• If x represents a count of the number of successes (the
number of occurrences of a certain event) then p = x/n
represents the proportion of successes in the sample.
• The sample proportion is approximately normally
distributed with the following parameters:
– The mean is the real proportion in the population: mp = p
– The standard deviation is: σ p  p(1 p)
n
Solve the problems
1.
A manufacturer of aspirin claims that the proportion of headaches
sufferers who get relief with just two aspirins is 53%.
•
•
2.
What is the probability in a random sample of 400 headache
sufferers, less than 50% obtain relief?
If indeed 50% of the sample actually obtained relief, what does this
suggest about the manufacturer’s claim?
A university bookstore claims that 50% of its customers are
satisfied with the service and price
•
•
If the claim is true, what is the probability that in a random sample of
600 customers, less than 45% are satisfied?
If 270 customers expressed satisfaction with the bookstore, what
does this tell you about the bookstore’s claim?
Solutions
1.
The probability to calculate is P(sample proportion<.5).
•
•
2.
To use the normal distribution we need to check two conditions:
np = (400)(.53) = 212> 5; np(1 – p) = 400(1-.53) = 188> 5.
Approximating the distribution of the sample proportion by the normal distribution should
lead to accurate results.
By the manufacturer’s statement:
m= .53. s=[p(1-p)/n]1/2 = [.53(1-.53)]1/2= .024955.
P(Proportion<.5)=P(Z<(.5 - .53)/.024955)=0.11465
The manufacturer claim is probably correct. The probability that the sample proportion is
less than 50% is only 11%, so if in a real sample the proportion turned up to be 50%
probably 53% in the whole population is a reasonable estimate.
The probability to calculate is P(Sample proportion < 45%).
•
•
[.5(1-.5)/600]1/2
Checking conditions: np = 600(.5) = 300 > 5, n(1-.5) = 300 > 5.
P(Proportion<.45) = normdist(.45,.50,.0204, true)=.007153
P(Proportion< 270/600) = P(Proportion<.45) = .007153 from the calculations above. If the
probability is only about 0.7% that the proportion of satisfied customers is less than 45%
(assuming the bookstore claim is true), and if actually only 45% of the customers are
satisfied in the real sample, the store claim is questionable.