# Statistics

of 17 powerpoint on statistics

## Content

Eco 045 Statistics
Lecture 7-1
Instructor: Suhui Li

•  Example
The average price of a 42-inch television on Best Buy’s
website is \$790. Assume that the price of these TVs follows
the normal distribution with a standard deviation of \$160.

Suppose that Ms. Li is shopping on Best Buy’s website for
a new 42-inch TV and has set a budget of between \$900
and \$1,000 for the purchase. There are 20 42-inch TVs on
the site. How many TVs are within her budget?

µ = 790

σ = 160

z1 = (x1 - µ)/σ
= (900 - 790)/160
= .69

z2 = (x2 - µ)/σ
= (1000 - 790)/160
= 1.31

P(900 < x < 1000)
= P(.69 < z < 1.31)
= P(z < 1.31) – P(z < .69)
= .9049 - .7549 = .15
15% of the TVs will fall into the price range of \$900 - \$1000
Thus, numver of TVs = 20*0.15 = 3

•  Example
A professor at a local community college noted that the
grades of his students were normally distributed with a
mean of 74 and a standard deviation of 10. The professor
has decided that 6.3 percent of his students received A's
while only 2.5 percent of his students failed the course and

1)  What is the minimum score needed to make an A?
2)  What is the maximum score among those who received
an F?

•  Example
A professor at a local community college noted that the
grades of his students were normally distributed with a
mean of 74 and a standard deviation of 10. The professor
has decided that 6.3 percent of his students received A's
while only 2.5 percent of his students failed the course and

1)  1-0.063 = 0.937 = P(z<1.53)
1.53=(x-74)/10 => x = 89.3
2)  0.025 = P(z<-1.96)
-1.96 = (x-74)/10 => x = 54.4

Normal Approximation of Binomial
Probabilities
•  A motivating example: Comcast Customer Service
An important part of the customer service of Comcast
relates to how fast the troubles in residential internet
service can be solved. Suppose that past data indicates that
the likelihood is 25% that such troubles can be solved in
one hour.
If the manager of customer service randomly select 5
cases, what is the probability that more than 3 troubles are
solved in one hour?

Normal Approximation of Binomial
Probabilities
What if the manager of customer service randomly select
48 cases, and is interested in the probability that more than
15 troubles are solved in one hour?
For n=5,
5
4
1
5
5
0
P(x>3) = P(x=4) + P(x=5) = C4 0.25 0.75 + C5 0.25 0.75
For n=48,
P(x>15) = P(x=16) + P(x=17) +…+ P(x=48)
= C1648 0.2516 0.7532 + C1748 0.25170.7531 +... + +C4848 0.25480.750

Normal Approximation of Binomial
Probabilities

Normal Approximation of Binomial
Probabilities
When the number of trials becomes large, the normal
probability distribution provides an easy-to-use
approximation of binomial probabilities
In the definition of the normal curve, set
µ = np and σ = np (1 − p)
The approximation generally improves as n increases.
As a rule of thumb, this normal approximation can be
used to when np > 5 and n(1 - p) > 5.

•  Example: Comcast Customer Service
Binomial probability: P(x>15) = P(x>16)
For n=48 and p=0.25
np > 5, n(1-p) > 5
μ = 12 and σ = 3
Normal approximation: P(x > 15.5)

15.5 − 12
z=
= 1.17
3

Not P(x>16)!

P(z>1.17) = 1- 0.8790 = 0.121
From Excel: true binomial probability P(x>15) = 0.1232

Normal Approximation of Binomial
Probabilities
Add and subtract a continuity correction factor
because a continuous distribution is being used to
approximate a discrete distribution.
Example:
P(x=8) ≈ P(7.5 < x < 8.5)
P(x>8) ≈ P(x > 7.5)
P(x<8) ≈ P(x < 8.5)

Normal Approximation of Binomial
Probabilities


Example
Suppose that a company has a history of making
errors in 10% of its invoices. A sample of 100
invoices has been taken, and we want to compute
the probability that 12 invoices contain errors.

Normal Approximation of Binomial
Probabilities


Example
In this case, we want to find the binomial
probability of 12 successes in 100 trials. So, we set:
µ = np = 100(.1) = 10
σ = np (1 − p) = [100(.1)(.9)] ½ = 3
P(x=12) = 0 for normal distribution!

Normal Approximation of Binomial
Probabilities


Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1

σ=3
P(11.5 < x < 12.5)
(Probability
of 12 Errors)

µ = 10
12.5
11.5

x

Normal Approximation of Binomial
Probabilities


Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1

P(x < 12.5) = .7967

10 12.5

x

Normal Approximation of Binomial
Probabilities


Normal Approximation to a Binomial Probability
Distribution with n = 100 and p = .1

P(x < 11.5) = .6915

10

x
11.5

Normal Approximation of Binomial
Probabilities


The Normal Approximation to the Probability
of 12 Successes in 100 Trials is .1052

P(x = 12)
= .7967 - .6915
= .1052

10

11.5

12.5

x

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