WARM-UPDetermine whether the following are linear transformations, combinations or both. Also find the new mean and standard deviation for the following.

1. A = 2.5x

2. B = X + Y

3. C = X – 2y

4. D = -.5y

5. E = xy

Mean Standard Deviation

X 16.5 2.5Y 20 4.5

COUNTING… Find the number of items in the sample space of

license plates containing 3 letters and 3 numbers that can be repeated.

What if they can’t be repeated?

PERMUTATIONS An arrangement of objects in a specific order

Order Matters and No Repetitions

EX: How many ways can you arrange 3 people in a picture?

EXAMPLE 2Suppose a business owner has a choice of 5 locations.

She decides the rank them from best to worst according to certain criteria. How many different ways can she rank them?

What if she only wanted to rank the top 3?

PERMUTATION RULE!

Where n = total # of objects and r = how many you need.

)!(

!

rn

nPrn

EXAMPLE 3A TV news director wishes to use 3 news stories on

the evening news. She wants the top 3 out of 8 possible. How many ways can the program be set up?

COMBINATIONSA selection of “n” objects without regard to order.

When different orderings of the same items are not counted separately we have a combination problem.EX: AB is the same as BA

When different ordering of the same items are counted separately, we have a permutation.EX: AB is different than BA

COMBINATION RULE

Example1 : To survey opinions of customers at local malls, a researcher decides to select 5 from 12. How many ways can this be done?

!)!(

!

rrn

nCrn

EXAMPLE 2 In a club, there are 7 women and 5 men. A committee of 3

women and 2 men is to be chosen. How many different possibilities are there?

What about a committee of 5 with at least 3 women?

At most 2 women?

BINOMIAL DISTRIBUTIONS Each trial has only 2 possible outcomes

“success” or “failure” There is a fixed # of trials (n) Trials are independent of each other The probability of a success (p) is constant

X ~ B(n, p) q – numerical probability of failure (1 – p) r – number of “successes” in n trials

BINOMIAL DISTRIBUTION FORMULA For X ~ B(n, p) then

xnx

rn

rnr

ppxnx

n

rnr

nC

r

n

qpr

nrXP

)1()!(!

!

)!(!

!

)(

EXAMPLE 1 A coin is tossed 3 times. Find the probability of getting

exactly 2 heads.

EXAMPLE 2 Public Opinion reported that 5% of Americans are afraid

of being alone in the house at night. If a random sample of 20 Americans is selected, find the probability that there are exactly 5 people who are afraid of being along in the house at night.

YOU TRY! A student takes a random guess at 5 multiple choice

questions. Find the probability that the student gets exactly 3 correct. Each question has 4 possible choices.

EXAMPLE 3 X is binomially distributed with 6 trials and a probability

of success equal to 1/5 at each.What is the probability of at least one success?

Three or fewer successes?

EXAMPLE 2 REVISITED Public Opinion reported that 5% of Americans are afraid

of being alone in the house at night. If a random sample of 20 Americans is selected.Find the probability that at most 3 are afraid.

Find the probability that at least 3 are afraid.

YOU TRY AGAIN! A student takes a random guess at 5 multiple choice

questions. Each question has 4 possible choices.Find the probability that the student gets at most 2

correct.

Find the probability that the student gets at least 2 correct.

MEAN & STANDARD DEVIATION For a binomial distribution:

p = probability of success and q = probability of failureμ = p and σ = √(pq) for 1 trialμ = 2p and σ = √(2pq) for 2 trialsμ = 3p and σ = √(3pq) for 3 trials

In general… μ = np and σ = √(npq) for n trials

EXAMPLE 1 5% of a batch of batteries are defective. A random sample

of 80 batteries is taken with replacement. Find the mean and standard deviation of the number of defective batteries in the sample.

2.

© 2011 P

earson Education, Inc

© 2011 P

earson Education, Inc

WARM UP A biased coin is tossed 6 times. The probability of

heads on any toss is 0.3. Let X denote the number of heads that come up.

Calculate:P(X = 2)P(X < 3)P(1 < X < 5).

NORMAL DISTRIBUTION A normal distribution curve is symmetrical,

bell-shaped curve defined by the mean and standard deviation of a data set.

The normal curve is a probability distribution with a total area under the curve of 1.

CHARACTERISTICS OF A NORMAL DISTRIBUTION

What do the 3 curves have in common?

CHARACTERISTICS OF A NORMAL DISTRIBUTION

The curves may have different mean and/or standard deviations but they all have the same characteristicsBell-shaped curveSymmetrical about the meanMean, median and mode are the same

(not skew!)

Area under the curve is always 1 (100%)

STANDARD NORMAL DISTRIBUTION Written as

Z ~ N(0, 1)Mean = 0 & Standard Deviation = 1

STANDARD NORMAL DISTRIBUTION Since the total area under the curve is 1, we can

consider partial areas to represent probabilities.

Z-SCORES A standard normal distribution is the set of

all z-scores. All values can be transformed

from a normal distribution toa standard normal by usingthe z-score.It represents how many standard

deviations “x” is always from the mean.The z-score is positive if the data value

lies above the mean and negative if the data value lies below the mean.

Z-SCORE EXAMPLES Suppose SAT scores among college students are

normally distributed with a mean of 500 and a standard deviation of 100. If a student scores a 700, what would be their z-score?

MORE Z-SCORE EXAMPLES For which test would a score of 78 have a higher

standing?A set of English test scores has a mean of 74 and a

standard deviation of 16.A set of math test scores has a mean of 70 and a

standard deviation of 8.

EVEN MORE Z-SCORE EXAMPLES What will be the miles per gallon for a Toyota Camry

when the average mpg is 23, it has a z-value of 1.5 and a standard deviation of 5?

AREA WITH A TABLEDraw the distribution curveShade the area in which you are interestedUse the table to find the areas

Might have to add or subtract to get what you want.

EXAMPLES FOR AREA Find the area/probability of the following:

Left of z = 1.99P(z < 1.99)

Left of z = 2.55P(z < 2.55)

Right of z = 1.11P(z > 1.11)

MORE EXAMPLES FOR AREA Find the area/probability of the following:

Left of z = -2.50P(z < -2.5)

Right of z = - 1.20P(z > -1.2)

EVEN MORE EXAMPLES FOR AREAFind the area/probability of the following:

P(0 < z < 2.32)

P(-1.2 < z < 2.3)

AND ONE MORE EXAMPLE FOR AREAFind the area/probability of the following:

P(z < -3.01 and z > 2.43)

APPLICATION 1 A Calculus exam is given to 500 students.

The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What percent of the students have scores between 82 and 90? How many students have scores between 82 and 90?

APPLICATION 2 A Calculus exam is given to 500 students.

The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What percent of the students have scores above 70? How many students scored above a 70?

APPLICATION 3 Find the probability of scoring below a

1400 on the SAT if the scores are normal distributed with a mean of 1500 and a standard deviation of 200.

FINDING Z-SCORES FROM AREA Find the z-score above the mean with an

area to the left of z equal to 0.9325

Find the z-score below the mean with an area to the left of z equal to 13.87%

MORE FINDING Z-SCORES FROM AREA Find the z-score below the mean with an

area between 0 and z equal to 0.4066

EVEN MORE FINDING Z-SCORES FROM AREA

Find the z-score above the mean with an area between 0 and z equal to 0.2123

Find the z to the right of the mean with an area to the right of z equal to 0.0239

INVERSE NORMAL DISTRIBUTIONS Find k for which P(x < k) = 0.95 given that x is normally

distributed with a mean of 70 and a standard deviation of 10.

APPLICATIONS A professor determines that 80% of this year’s History

candidates should pass the final exam. The results are expected to be normally distributed with a mean of 62 and standard deviation of 13. Find the lowest score necessary to pass the exam.

MORE APPLICATIONS Researchers want to select people in the middle 60% of

the population based on their blood pressure. If the mean is 120 and the S.D. is 8. Find the upper and lower reading that would qualify.

FINDING STATS BASED ON PROBABILITY

Sacks of potatoes with a mean weight of 5 kg are packed by an automatic loader. In a test, it was found that 10% of bags were over 5.2 kg. Use this information to find the standard deviation of the process

MORE FINDING STATS BASED ON PROBABILITY

Find the mean and the standard deviation of a normally distributed random variables X, if P(x > 50) = 0.2 and P(x < 20) = 0.3