NORMAL DISTRIBUTION INVERSE NORMAL Z-CURVE BINOMIAL DISTRIBUTION NORMAL APPROXIMATION TO BINOMIAL...

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NORMAL DISTRIBUTIONINVERSE NORMAL

Z-CURVEBINOMIAL DISTRIBUTION

NORMAL APPROXIMATION TO BINOMIAL

Chapter 4 Binomial and Normal Distributions

Normal Distribution: N(mean,var)

Symmetric about the mean

The mean divides the curve in 2 halves

The probability is the area under the curve

The total area or probability under the curve is 1

It is completely determined by the mean and standard deviation

Normal Distribution

They sell an average of 36 TVs with standard deviation of 8. How many TVs do they need to have to meet the demand? Compute the probability of selling 40 or more TV sets.

Histogram and stemplot of sales data of large screen TVs.

Normal Distribution

Computing Probabilities using TI2nd Distribution Normalcdf

Normalcdf(a,b,mean,standard deviation)

Normal Distribution

Sales of large screen TVs exampleMean = 36 SD = 8

Compute the probability of selling 40or more TV sets.

P(X≥40) = Normalcdf(40,9999,36,8) = 0.3085

Normal Distribution

P(a≤X≤b)=normalcdf(a,b,mean,sd)

e.g., Mean = 36 SD = 8

P(40≤X≤50)=normalcdf(40,50,36,8)

=0.2685

Normal Distribution

P(X≥a)=normalcdf(a,99999,mean,sd)

e.g., Mean = 36 SD = 8

P(X≥40)=normalcdf(40,99999,36,8) =0.3085

Normal Distribution

P(X≤a)=normalcdf(-99999, a,mean sd)

e.g., Mean = 36 SD = 8

P(X≤40)=normalcdf(-99999, 40, 36, 8)=0.6915

iClicker

A study was conducted to determine if any link existed between cellular phone usage and brain cancer development. Data from this study indicated that the monthly phone usage for all users is approximately normally distributed with mean 2.4 hours and standard deviation 1.1 hours. What is the proportion of phone users who use their phones at least 3 hours a month?a. 0.7073b. 0.2143c. -0.2927d. 0.6927e. 0.2927

Inverse Normal

a = ? Such that P(X≤a) = Percentile

InvNorm(percentile, mean, sd)

2nd distribution InvNorm

Note: InvNorm computes a so that the area from the left most to a is the percentile

Inverse Normal

From the TV example, how many TVs do theyneed in stock to meet the demand? The store manager decided to order enough TVs to cover 90% of the demand. How many TVs should they order?

a = number of TVs they should orderP(X≤a) = 0.90a = InvNorm(.90,36,8)a = 46.25 ≈ 46 TVs

Inverse Normal

Sales and Demand of TVs exampleMean = 36 SD = 8

What is a when P(X ≤ a) = 0.70a = InvNorm(.70, 36, 8) = 40.20

What is a when P(X ≤ a) = 0.30a = InvNorma(.30, 36, 8) = 31.80

What is a when P(X ≥ a) = 0.60P(X ≤ a) = 0.40a = InvNorm(.40, 36, 8) = 33.97

iClicker

A study was conducted in order to determine if any link existed between cellular phone usage and the development of brain cancer. Data from this study indicated that the monthly phone usage for all users is approximately normally distributed with mean 2.4 hours and standard deviation 1.1 hours. Up to how many hours does the lower 80% of the users spend on the phone?a. 1.82 hoursb. 3.33 hoursc. 1.47 hoursd. 3.12 hourse. 3.81 hours

Z-Score

Z is how many standard units above or below the mean (standard deviations from the mean)

Z = (X – Mean)/ SDProperties of Z curve

Z Curve is Normal with Mean = 0 and SD = 1

P ( -1 ≤ Z ≤ 1 ) = .68P ( -2 ≤ Z ≤ 2 ) = .95P ( -3 ≤ Z ≤ 3 ) = .997

For a certain process, the chance of observing at most 10 cases is 5%. The process is known to have a standard deviation of 5. What is the mean of the process?

)1,0,05.0(*510

)1,0,05.0(5

05.0)10(

InvNorm

Z-Score

Binomial Distribution

Properties of a Binomial Process trial performed n times and each trial is independent of each

other each trial results in one of two possible outcomes (S = Success, F

= Failure) the probability of success is p and the probability of failure is q = 1 – p for all trials

Binomial Random Variable

X = total number of successes

Parameters: n = number of trials p = probability of success

Example 1A Stat 216 multiple choice quiz has 5

questions, with each question having 4 choices.

a. Success = the question answered correctlyb. X = number of correct answers c. X = {0,1,2,3,4,5}d. n = 5e. p = ¼

Example 2Available data shows that 40% of telephone

respondents agree to be interviewed for market research surveys. Suppose that the polling organization Reliable Research randomly selects and dials telephone numbers until 50 respondents are reached.

a. Success = respondent agrees to be interviewedb. X = number of respondents who agree to be interviewedc. X = {0,1,2,3,…,49,50}d. n = 50e. p = 0.40

Calculator. P(X = j) = binompdf(n, p, j) 2nd distribution binompdfExample1. Calculate the probability of each

event occurring. (n=5 , p=0.25)

Note: The total probability should be 1.0P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5) = 1

P(X=0) P(X=1) P(X=2) P(X=3) P(X=4) P(X=5)binompdf(5,0.25,0) binompdf(5,0.25,0) binompdf(5,0.25,0) binompdf(5,0.25,0) binompdf(5,0.25,0) binompdf(5,0.25,0)

0.237 0.396 0.264 0.088 0.015 0.0009

iClicker

Calculator: P(X ≤ j ) = binomcdf(n,p,j)P(X ≤ j ) = P(X = 0) + P(X = 1) + . . . + P(X = j)From Example1, what is the probability that a

student will get at most 3 correct answers?

984375.0

)3,25.0,5(

)3()2()1()0()3(

binomcdf

XPXPXPXPXP

iClicker

From Example1, what is the probability that a student will get at least 2 correct answers?

3671875.0

)1,25.0,5(1

)1()0(1

)5()4()3()2()2(

binomcdf

XPXPXPXPXP

iClicker

If X is defined to be a binomial random variable with parameters n and p then:

the expected value or the mean of X is np the standard deviation of X is )1( pnp

From Example2, how many respondents will be expected to agree to be interviewed? What is the corresponding mean and standard deviation?

n = 50 p = .40

)40.1(*40.*50

40.*50

npmean The researcher should expect to interview around 20 respondents give or take 4.

Binominal Distribution with n=30, p=0.10

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Binomial Distribution with n=30, p=0.85

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Binomial Distribution with n=30, p=0.45

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

Right-skewed

Left-skewed

Symmetric

Normal Approximation to Binomial

The normal curve gives reasonably good approximations of binomial probabilities whenever both np > 5 and n(1-p) > 5.

To which case above will the normal approximation to the binomial be valid?

np n(1-p)n = 30, p = 0.10 3 27n = 30, p = 0.85 25.5 4.5n = 30, p = 0.45 13.5 16.5

NORMAL DISTRIBUTION INVERSE NORMAL Z-CURVE BINOMIAL DISTRIBUTION NORMAL APPROXIMATION TO BINOMIAL...

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