1

Counting principles, including permutations and combinations.

The binomial theorem: expansion of 𝒂 + 𝒃 𝒏, 𝒏 𝜺 𝑵.

● THE PRODUCT RULE

If there are 𝑚 different ways of performing an operation and for each of these there are 𝑛 different ways of performing a second independent operation, then there are 𝑚𝑛 different ways of performing the two operations in succession. The product principle can be extended to three or more successive operations. The number of different ways of performing an operation is equal to the sum of the different mutually exclusive possibilities.

● COUNTING PATHS

The word 𝑎𝑛𝑑 suggests multiplying the possibilities The word 𝑜𝑟 suggests adding the possibilities. If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation.

● PERMUTATIONS (order matters)

A permutation of a group of symbols is any arrangement of those symbols in a definite order.

● Permutations of 𝒏 different object : 𝒏!

Explanation: Assume you have n different symbols and therefore n places to fill in your arrangement. For the first place, there are n different possibilities. For the second place, there are n – 1 possible symbols, … until we saturate all the places. According to the product principle, therefore, we have n (n – 1)(n – 2)(n – 3)⋯1 different arrangements, or n!

Wise Advice: If a group of items have to be kept together, treat the items as one object. Remember that there may

be permutations of the items within this group too. ● Permutations of 𝒌 different objects out of 𝒏 different available (no repetition allowed) :

𝑃𝑘𝑛 =

𝑛!

𝑛 − 𝑘 != 𝑛 ∙ 𝑛 − 1 ∙∙∙ 𝑛 − 𝑘 + 1

Good logic to apply to similar questions straightforward: Suppose we have 10 letters and want to make groups of 4 letters. For four-letter permutations, there are 10 possibilities for the first letter, 9 for the second, 8 for the third, and 7 for the last letter. We can find the total number of different four-letter permutations by multiplying 10 x 9 x 8 x 7 = 5040.

● Permutations with repetition of 𝒌 different objects out of 𝒏 different available = 𝒏𝒌

There are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multiplying each time.)

● COMBINATIONS (order doesn’t matters)

It is the number of ways of choosing 𝒌 objects out of 𝒏 available given that ▪ The order of the elements does not matter. ▪ The elements are not repeated [such as lottery numbers (2,14,15,27,30,33)]

The easiest way to explain it is to:

assume that the order does matter (ie permutations),

then alter it so the order does not matter. Since the combination does not take into account the order, we have to divide the permutation of the total number of symbols available by the number of redundant possibilities. 𝒌 selected objects have a number of redundancies equal to the permutation of the objects 𝒌! (since order doesn’t matter)

2

However, we also need to divide the permutation n! by the permutation of the objects that are not selected, that is to say 𝒏 − 𝒌 ! .

𝒏!

𝒌! 𝒏 − 𝒌 !

(𝒏𝒌) ≡ 𝑪𝒌

𝒏 ≡ 𝑪𝒌𝒏 =

𝒏!

𝒌! 𝒏 − 𝒌 !=

𝒏 𝒏 − 𝟏 𝒏 − 𝟐 𝒏 − 𝒌 + 𝟏

𝒌!

● Binomial Expansion/Theorem

𝑎 + 𝑏 𝑛 = ∑ (𝑛𝑘) 𝑎𝑛−𝑘𝑏𝑘 = 𝑎𝑛 + (

𝑛1) 𝑎𝑛−1𝑏 + ⋯+ (

𝑛𝑘) 𝑎𝑛−𝑘𝑏𝑘 + ⋯+ 𝑏𝑛

𝑛

𝑘=0

● Binomial Coefficient

(𝑛𝑘) is the coefficient of the term containing 𝑎𝑛−𝑘𝑏𝑘 in the expansion of 𝑎 + 𝑏 𝑛

(𝑛𝑘) =

𝑛 𝑛 − 1 𝑛 − 2 ⋯ 𝑛 − 𝑘 + 1

𝑘!=

𝑛!

𝑘! 𝑛 − 𝑘 !=

𝑛!

𝑛 − 𝑘 ! 𝑘!= (

𝑛𝑛 − 𝑘

)

𝑇ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑡𝑒𝑟𝑚 𝑜𝑟 𝑘 + 1 𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠: 𝑇𝑘+1 = (𝑛𝑘)𝑎𝑛−𝑘𝑏𝑘

The constant term is the term containing no variables.

When finding the coefficient of 𝑥𝑛 always consider the set of all terms containing 𝑥𝑛

Probability The number of trials is the total number of times the “experiment” is repeated.

The outcomes are the different results possible for one trial of the experiment.

Equally likely outcomes are expected to have equal frequencies.

The sample space, U, is the set of all possible outcomes of an experiment.

And event is the occurrence of one particular outcome.

𝑃 𝐴 =𝑛 𝐴

𝑛 𝑈

Complementary Events

Two events are described as complementary if they are the only two possible outcomes.

Two complementary events are mutually exclusive.

Since an event must either occur or not occur, the probability of the event either occurring or not

occurring must be 1.

𝑷 𝑨 + 𝑷 𝑨′ = 𝟏

Use when you need probability that an event will not happen

Possibility when we are interested in more than one outcome (events are “and”, “or”, “at least”)

P(A) is the probability of an event A occurring in one trial,

n(A) is the number of times event A occurs in the sample space n(U) is

the total number of possible outcomes.

𝑛 𝜀 𝑁 (𝑛0) ≡ 1 0! ≡ 1

3

Combined Events

∪ 𝑢𝑛𝑖𝑜𝑛 ≡ 𝑒𝑖𝑡ℎ𝑒𝑟 ∩ 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛 ≡ 𝑏𝑜𝑡ℎ/𝑎𝑛𝑑

Given two events, B and A, the probability of at least one of the two events occurring,

𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵

𝐼𝑡 𝑖𝑠 𝑖𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑡 𝑡𝑜 𝑘𝑛𝑜𝑤 ℎ𝑜𝑤 𝑡𝑜 𝑔𝑒𝑡 𝑃 𝐴 ∩ 𝐵

For mutually exclusive events (no possibility that A and B occurring at the same time)

Turning left and turning right (you can't do both at the same time)

Tossing a coin: Heads and Tails

𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 𝑃 𝐴 ∩ 𝐵 = ∅

For non - mutually exclusive we are going to find conditional probability for

Independent and Dependent Events

A bag contains three different kinds of marbles: red, blue and green. You pick the marble twice. Probability of picking up the red one (or any) the second time depends weather you put back the first marble or not.

• Independent Events: • Dependent Events:

the probability that one event occurs probability of one event occurring influences in no way affects the probability of the likelihood of the other event the other event occurring. You put the first marble back You don’t put the first marble

∎ Conditional Probability:

Given two events, B and A, the conditional probability of an event A is the probability that the event will occur given the knowledge that an event B has already occurred. This probability is written as (notation for the probability of A given B) P (A|B )

Probability of the intersection of A and B (both events occur) is: 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐵 𝑃 𝐴|𝐵

• Independent Events: • Dependent Events:

𝑃 𝐴|𝐵 = 𝑃 𝐴 = 𝑃 𝐴|𝐵′ 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐵 𝑃 𝐴|𝐵

𝐴 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑑𝑒𝑝𝑒𝑛𝑑 𝑜𝑛 𝐵 𝑛𝑜𝑟 𝑜𝑛 𝐵′ 𝑃 𝐴|𝐵 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡 𝐵

𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃 𝐵 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐵 𝑃 𝐴|𝐵

𝑃 𝐴|𝐵 =𝑃 𝐴 ∩ 𝐵

𝑃 𝐵

either A or B or both

P(A) includes part of B from intersection P(B) includes part of A from intersection

𝑃 𝐴 ∩ 𝐵 (both A and B) was counted twice, so one has to be subtracted

4

● Use of Venn diagrams, tree diagrams and tables of outcomes to solve problems.

1. Venn Diagrams

The probability is found using the principle 𝑃 𝐴 =𝑛 𝐴

𝑛 𝑈

2. Tree diagrams

A more flexible method for finding probabilities is known as a tree diagram.

⧪ Bayes’ Theorem

𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐵 𝑃 𝐴|𝐵 ⟹ 𝑃 𝐴 ∩ 𝐵 = 𝑃 𝐴 𝑃 𝐵|𝐴

𝑃 𝐴|𝐵 = 𝑃 𝐴 ∩ 𝐵

𝑃 𝐵 =

𝑃 𝐴 𝑃 𝐵|𝐴

𝑃 𝐵 𝐵𝑎𝑦𝑒𝑠′ 𝑡ℎ𝑒𝑜𝑟𝑒𝑚

▪ Another form of Bayes’ theorem (Formula booklet)

From tree diagram:

there are two ways to get A, either after B has happen or after B has not happened:

𝑃 𝐴 = 𝑃 𝐵 𝑃 𝐴|𝐵 + 𝑃 𝐵′ 𝑃 𝐴|𝐵′ ⟹ 𝑃 𝐵|𝐴 =𝑃 𝐵 𝑃 𝐴|𝐵

𝑃 𝐵 𝑃 𝐴|𝐵 + 𝑃 𝐵′ 𝑃 𝐴|𝐵′

This allows one to calculate the probabilities

of the occurrence of events, even where trials

are non-identical (where 𝑃 𝐴|𝐴 ≠ 𝑃 𝐴 ),

through the product principle.

5

▪ Extension of Bayes’ Theorem

If there are more options than simply B occurs or B doesn’t occur, for example if there were three possible outcomes for the first event B1, B2, and B3

Probability of A occurring is: 𝑃 𝐵1 𝑃 𝐴|𝐵1 + 𝑃 𝐵2 𝑃 𝐴|𝐵2 + 𝑃 𝐵3 𝑃 𝐴|𝐵3

𝑃 𝐵𝑖|𝐴 =𝑃 𝐵𝑖 𝑃 𝐴|𝐵𝑖

𝑃 𝐵1 𝑃 𝐴|𝐵1 + 𝑃 𝐵2 𝑃 𝐴|𝐵2 + 𝑃 𝐵3 𝑃 𝐴|𝐵3

Outcomes B1, B2, and B3 must cover all the possible outcomes.

Descriptive Statistics

Concepts of population, sample, random sample and frequency distribution of discrete and continuous data.

A population is the set of all individuals with a given value for a variable associated with them.

A sample is a small group of individuals randomly selected (in the case of a random sample) from the population as a whole, used as a representation of the population as a whole.

The frequency distribution of data is the number of individuals within a sample or population for each value of the associated variable in discrete data, or for each range of values for the associated variable in continuous data.

new guidelines in IB MATH: population ≡ sample

Presentation of data: frequency tables and diagrams Grouped data: mid-interval values, interval width, upper and lower interval boundaries, frequency histograms.

Mid interval values are found by halving the difference between the upper and lower interval boundaries.

The interval width is simply the distance between the upper and lower interval boundaries.

Frequency histograms are drawn with interval width proportional to bar width and frequency as the height.

Median, mode; quartiles, percentiles.

Range; interquartile range; variance, standard deviation.

Mode (discrete data) is the most frequently occurring value in the data set.

Modal class (continuous data) is the most frequently occurring class.

Median is the middle value of an ordered data set.

For an odd number of data, the median is middle data.

For an even number of data, the median is average of two middle data.

6

Percentile is the score bellow which a certain percentage of the data lies.

Lower quartile (Q1) is the 25th percentile.

Median (Q2) is the 50th percentile.

Upper quartile (Q3) is the 75th percentile.

Range is the difference between the highest and lowest value in the data set.

The interquartile range is Q3−Q1.

Cumulative frequency is the frequency of all values less than a given value.

The population mean, μ is generally unknown but the sample mean, �̅� used to serve as an unbiased estimate of this mean. That used to be. From now on for the examination purposes, data will be treated as the population. Estimation of mean and variance of population from a sample is no longer required.

Discrete and Continuous Random Variables

A variable X whose value depends on the outcome of a random process is called a random variable. For any random variable there is a probability distribution/ function associated with it.

● Probability distribution/ function

Discrete Random Variables

P(X = x), the probability distribution of x, involves listing P(𝑥𝑖 ) for each 𝑥𝑖 .

1. 0 ≤ 𝑃 𝑋 = 𝑥 ≤ 1

2. ∑𝑃 𝑋 = 𝑥 = 1

𝑥

3. 𝑃 𝑋 = 𝑥𝑛 = 1 − ∑ 𝑃 𝑋 = 𝑥𝑘

𝑘≠𝑛

[𝑃 𝑒𝑣𝑒𝑛𝑡 𝑥𝑛 𝑜𝑐𝑐𝑢𝑟𝑠 = 1 − 𝑃 𝑎𝑛𝑦 𝑜𝑡ℎ𝑒𝑟 𝑒𝑣𝑒𝑛𝑡 𝑜𝑐𝑐𝑢𝑟𝑠 ]

Median given by middle term; if even number of terms, average of 2 middle terms.

Same applies for Q1, Q2 and IQR.

It is the value of X such that 𝑃 𝑋 ≤ 𝑥 ≥1

2 𝑎𝑛𝑑 𝑃 𝑋 ≥ 𝑥 ≥

1

2

The mode is the value of x with largest 𝑃 𝑋 = 𝑥 which can be different from the expected value

ALWAYS watch out for conditional probability (i.e. P(x) given that); it is often implied and not stated

𝐸 𝑋 = 𝜇 = 𝑥 𝑃 𝑋 = 𝑥 ≡ 𝑝𝑖𝑥𝑖 =

𝑓𝑖𝑥𝑖

𝑓𝑖

𝐼𝑓 𝑎 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑡ℎ𝑒𝑛 𝐸 𝑎𝑋 = 𝑎𝐸 𝑋

𝐼𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠, 𝑡ℎ𝑒𝑛 𝐸 𝑎𝑋 + 𝑏 = 𝑎𝐸 𝑋 + 𝑏

𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌

𝐸[ 𝑔 𝑋 ] = 𝑔 𝑥 𝑃 𝑋 = 𝑥

𝑉𝑎𝑟 𝑋 = 𝜎2 = 𝐸 𝑋 – 𝜇 2 =𝑓

𝑖 𝑥𝑖 − 𝜇 2

𝑓𝑖

𝜎2 = 𝐸 𝑋2 − 𝜇2

𝑉𝑎𝑟 𝑎𝑋 + 𝑏 = 𝑎2𝑉𝑎𝑟 𝑋

𝑉𝑎𝑟[𝑋 + 𝑌] = 𝑉𝑎𝑟 𝑋 + 𝑉𝑎𝑟 𝑌 true only for independent

7

Continuous Random Variables X defined on a ≤ x ≤ b

probability density function (p.d.f.), f (x), describes the relative likelihood for this variable to take on a given value cumulative distribution function c.d.f.), 𝐹 𝑡 , is found by integrating the p.d.f. from minimum

value of X to t

𝐹 𝑡 = 𝑃 𝑋 ≤ 𝑡 = ∫ 𝑓 𝑥 𝑑𝑥𝑡

𝑎

For a function 𝑓 𝑥 to be probability function, it must satisfy the following conditions:

1. 𝑓 𝑥 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝜖 𝑎, 𝑏

2. ∫ 𝑓 𝑥 = 1𝑏

𝑎

3. 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑎 ≤ 𝑐 < 𝑑 ≤ 𝑏, 𝑃 𝑐 < 𝑋 < 𝑑 = ∫ 𝑓 𝑥 𝑑𝑥𝑑

𝑐

▪ 𝐹𝑜𝑟 𝑎 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒, 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑛𝑦 𝑠𝑖𝑛𝑔𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 𝑧𝑒𝑟𝑜

𝑃 𝑋 = 𝑐 = 0 ⇒ 𝑃 𝑐 ≤ 𝑋 ≤ 𝑑 = 𝑃 𝑐 < 𝑋 < 𝑑 = 𝑃 𝑐 ≤ 𝑋 < 𝑑 𝑒𝑡𝑐.

Median a number m such that ∫ 𝑓 𝑥 = 1/2𝑚

𝑎

Mode: max on 𝒇 𝒙 , 𝑎 < 𝑥 < 𝑏 (which may not be unique).

ALWAYS watch out for conditional probability (i.e. P(x) given that); it is often implied and not stated

𝐸 𝑋 = 𝜇 = ∫ 𝑥 𝑓 𝑥 𝑑𝑥 𝐼𝑓 𝑎 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑡ℎ𝑒𝑛 𝐸 𝑎𝑋 = 𝑎𝐸 𝑋

𝐼𝑓 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠, 𝑡ℎ𝑒𝑛 𝐸 𝑎𝑋 + 𝑏 = 𝑎𝐸 𝑋 + 𝑏

𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌

𝐸[ 𝑔 𝑋 ] = 𝑔 𝑥 𝑃 𝑋 = 𝑥

𝑉𝑎𝑟 𝑋 = 𝜎2 = ∫ 𝑥2 𝑓 𝑥 𝑑𝑥 − [∫ 𝑥𝑓 𝑥 𝑑𝑥𝑏

𝑎]2𝑏

𝑎

𝜎2 = 𝐸 𝑋2 − 𝜇2

𝑉𝑎𝑟 𝑎𝑋 + 𝑏 = 𝑎2𝑉𝑎𝑟 𝑋

𝑉𝑎𝑟[𝑋 + 𝑌] = 𝑉𝑎𝑟 𝑋 + 𝑉𝑎𝑟 𝑌 true only for independent Standard deviation of X: 𝜎 = √𝑉𝑎𝑟 𝑋

CALCULATOR

Binomial Distribution • 𝑿 ~ 𝑩(𝒏, 𝒑)

• n is number of trials • There is either success or failure

• p is the probability of a success

• (1 – p) is the probability of a failure.

• 𝑃(𝑋 = 𝑥) = (𝑛𝑥

) 𝑝𝑥(1 − 𝑝)𝑛−𝑥 𝑥 = 0, 1, … , 𝑛

• 𝐸(𝑋) = 𝜇 = 𝑛𝑝

• 𝑉𝑎𝑟(𝑋) = 𝜎2 = 𝑛𝑝(1 − 𝑝)

In a given problem you write: 𝑋 ~ 𝐵(100, 0.5) 𝑃(𝑥 ≤ 52) = 0.6913502844

Poisson Distribution • 𝑿 ~ 𝑷𝒐(𝒎)

The average/mean number of occurrences (m) is constant for every interval. The probability of more than one occurrence in a given interval is very small. The number of occurrences in disjoint intervals are independent of each other.

• 𝑃(𝑋 = 𝑥) =𝑚𝑥𝑒−𝑚

𝑥!, 𝑥 = 0, 1, 2, …

• 𝐸(𝑋) = 𝑚

• 𝑉𝑎𝑟(𝑋) = 𝑚

In a given problem you write: 𝑋 ~ 𝑃𝑜(0.325) 𝑃(𝑥 ≥ 6) = 1 − 𝑃(𝑥 ≤ 5) 𝑃(𝑥 ≥ 6) = .3840393444

If 𝑋 ~ 𝑃𝑜(ℓ), and 𝑌 ~ 𝑃𝑜(𝑚), 𝑋 + 𝑌~ 𝑃𝑜(ℓ + 𝑚)

BinomCDF(trials , probability of event , value)

o Gives cumulative probability, i.e. the number of successes within n trials is at most the value

𝑎𝑡 𝑚𝑜𝑠𝑡: 𝑃(𝑋 ≤ 𝑥) = 𝑏𝑖𝑛𝑜𝑚𝑐𝑑𝑓(𝑛, 𝑝, 𝑥)

𝑎𝑡 𝑙𝑒𝑎𝑠𝑡:

𝑃(𝑋 ≥ 𝑥) = 1 − 𝑏𝑖𝑛𝑜𝑚𝑐𝑑𝑓 (𝑛, 𝑝, 𝑥 − 1).

PoissonCDF(mean , value) o Gives cumulative probability, i.e. probability of at most (value) occurrences within a time period

𝑎𝑡 𝑚𝑜𝑠𝑡: 𝑃(𝑋 ≤ 𝑥) = 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓(𝑚, 𝑥) 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡:

𝑃(𝑋 ≥ 𝑥) = 1 − 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑐𝑑𝑓 (𝑚, 𝑥 − 1).

BinomPDF(trials , probability of event , value)

o Gives the probability for a particular number of success in n trials

𝑒𝑥𝑎𝑐𝑡𝑙𝑦: 𝑃(𝑋 = 𝑥 ) = 𝑏𝑖𝑛𝑜𝑚𝑝𝑑𝑓 (𝑛, 𝑝, 𝑥)

PoissonPDF(mean , value) o Gives probability of a particular number of occurrences within a time period

𝑒𝑥𝑎𝑐𝑡𝑙𝑦: 𝑃(𝑋 = 𝑥 ) = 𝑝𝑜𝑖𝑠𝑠𝑜𝑛𝑝𝑑𝑓 (𝑚, 𝑥)

Normal distribution. • 𝑿 ~ 𝑵(𝝁, 𝝈𝟐)

• 𝑓(𝑥) =1

𝜎√2𝜋 𝑒−

12

(𝑥−𝜇

𝜎)

2

− ∞ < 𝑥 < ∞

• ∫ 𝑓(𝑥)𝑑𝑥 = 1∞

−∞

• 𝑃(𝑥 ≤ 𝑥1) = 𝑎𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥𝑥1

−∞

• 𝑃(𝑥 ≥ 𝑥2) = 𝑎𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥∞

𝑥2

• 𝑃(𝑥1 ≤ 𝑥 ≤ 𝑥2) = 𝑎𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥𝑥2

𝑥1

• 𝑃(𝑋 = 𝑥1) = 0 ⇒ 𝑃(𝑥1 ≤ 𝑋 ≤ 𝑥2) = 𝑃(𝑥1 < 𝑋 < 𝑥2) = 𝑃(𝑥1 ≤ 𝑋 < 𝑥2) 𝑒𝑡𝑐.

In a given problem you write: 𝜇 = 70 SD = 4.5 𝑋 ~ 𝑁(70, 20.25) 𝑃(65 ≤ 𝑋 ≤ 80) = 0.8536055925

probability is 85.4%.

In a given problem you write: 𝜇 = 2870; 𝜎 = 900 𝑃(𝑥 ≤ 𝑎) = 0.3409

𝑎 = 2500

Standardized normal distribution • 𝒁 ~ 𝑵(𝟎, 𝟏)

𝑍 =𝑋 − 𝜇

𝜎 (z − score ) 𝑋 = 𝜇 + 𝑍𝜎 𝑖𝑡 𝑡𝑒𝑙𝑙𝑠 𝑢𝑠 𝑤ℎ𝑒𝑟𝑒 𝑖𝑠 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑋 𝑖𝑛 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝜎

invNorm(probability) o Given a probability, gives the corresponding z-score.

𝑃(𝑧 ≤ 𝑏) = Ф(𝑏)

𝑎 = 𝐼𝑛𝑣𝑁𝑜𝑟𝑚(Ф(𝑏))

𝑃(𝑍 ≤ 𝑏) = 𝑃(𝑍 < 𝑏) 𝑏𝑐 𝑃(𝑍 = 𝑏) = 0

NormalCDF(lower , upper , mean , SD) o Gives probability that a value is within a given range

o percentage of area under a continuous distribution curve from lower bound to upper bound.

𝑃(𝑥1 ≤ 𝑋 ≤ 𝑥2) = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 ( 𝑥1, 𝑥2, 𝜇, 𝜎)

𝑃(𝑥 ≤ 𝑥) = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 (−1𝐸99, 𝑥, 𝜇, 𝜎)

𝑃(𝑋 ≥ 𝑥 ) = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 ( 𝑥, 1𝐸99, 𝜇, 𝜎)

InvNorm(probability, 𝝁, 𝝈) o Given the probability, this function returns the x – value region to the left of x – value.

𝑃(𝑥 ≤ 𝑎) = Ф(𝑎) = 𝑎𝑟𝑒𝑎 (𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦)

𝑎 = 𝐼𝑛𝑣𝑁𝑜𝑟𝑚(𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦, 𝜇, 𝜎)

NormalCDF(lower , upper) o Gives probability that a value is within a given range

𝑃(𝑧1 ≤ 𝑍 ≤ 𝑧2) = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 ( 𝑧1, 𝑧2)

𝑃(𝑍 ≤ 𝑧 ) = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 (−1𝐸99, 𝑧)

𝑃(𝑍 ≥ 𝑧 ) = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 (𝑧, 1𝐸99)

8