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

received F's.

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

received F's.

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

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

received F's.

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

received F's.

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