Week 15 - Rules of probability, and probability distributions Pt3

Question 1

What is a random variable?

Answer 1

A random variable is a numerical description of the outcome of an
experiment

Question 2

What is a discrete random variable?

Answer 2

A discrete random variable may assume either a finite number of
values or an infinite sequence of values

Question 3

What is a continuous random variable?

Answer 3

A continuous random variable may assume any numerical value in an
interval or collection of intervals.

Question 4

What are the 4 properties of a binomial experiment?

Answer 4

• The experiment consists of a sequence of n identical trials.
• Two outcomes, success and failure, are possible on each trial.
• The probability of a success, denoted by p, does not change from trial to trial.
• The trials are independent.

Question 5

What is the interest in binomial distribution?

Answer 5

Our interest is in the number of successes occurring in the n trials.

We let X denote the number of successes occurring in the n trials

Question 6

What does the binomial probability function look like?

Answer 6

𝑝(𝑥)= 𝑛!/(𝑥!(𝑛−𝑥)!) [𝜋^𝑥 (1−𝜋)^(𝑛−𝑥)]

Where:
p(x) = the probability of x success in n trials
n = the number of trials
𝜋= the probability of success on any one trial

Question 7

What does this portion of the binomial probability function mean?
𝑛!/(𝑥!(𝑛−𝑥)!)

Answer 7

number of experimental outcomes providing exactly x successes in n trials

Question 8

What does this portion of the binomial probability function mean?
𝜋^𝑥 (1−𝜋)^(𝑛−𝑥)

Answer 8

Number of experimental
outcomes providing exactly
x successes in n trials

Question 9

What is the formula for the expected value (mean) of a binomial distribution?

Answer 9

𝐸(𝑋)= 𝜇=𝑛𝜋

Question 10

What does the expected value in a binomial distribution represent?

Answer 10

The average number of successes in
n trials with success probability π.

Question 11

What is the formula for the variance of a binomial distribution?

Answer 11

𝜎^2= 𝑛𝜋(1−𝜋)

Question 12

What does the variance in a binomial distribution measure?

Answer 12

The variability in the number of successes.

Question 13

What is the formula for the standard deviation of a binomial distribution?

Answer 13

𝜎= √(𝑛𝜋(1−𝜋) )

Question 14

What does the standard deviation in a binomial distribution represent?

Answer 14

The spread of the number of successes around the mean, in the same units as
X.

Question 15

Example
Evian is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually.
Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.

Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year?

Answer 15

Using the Binomial Probability Function
p(x) = the probability of x success in n trials = 0.10
n = the number of trials = 3
𝑥= number of success = 1

𝑝(𝑥)= 𝑛!/(𝑥!(𝑛−𝑥)!) [𝜋^𝑥 (1−𝜋)]^(𝑛−𝑥)

𝑝(𝑥)= 3!/(1!(3−1)!) (0.10)^1 (1−0.10)^(3−1) = 0.243

Expected Value = 𝐸(𝑋)= 𝜇=𝑛𝜋 = 3*0.1 = 0.3 employees out of 3

Variance = 𝜎^2= 𝑛𝜋(1−𝜋) = 3(0.1)(0.9) = 0.27

Standard Deviation = 𝜎= √(𝑛𝜋(1−𝜋) ) = 0.52 employees

Question 16

Example
A broker has a bonus scheme to encourage profitable trading. Under the rules of the scheme, any trader who drops below his daily target more than three times in a two-week period (10 working days) will forfeit his bonus at the end of the period. If the probability that an employee will be below target on any one day is 0.15, how many bonuses will be lost by 100 traders in a 50-week year? (Assumptions of independence are valid here).
a. 50
b. 100
c. 125
d. 10

Answer 16

P(X>3) = 1 – P(X=0)- P(X=1) - P(X=2) - P(X=3)

= 1 − 𝐶10_0∗ 0.15^0∗ 0.85^10 − 𝐶10_1∗ 0.15^1 ∗ 0.85^9 − 𝐶10_2 ∗ 0.15^2 ∗ 0.85^8 − 𝐶10_2 ∗ 0.15^3 ∗ 0.85^7

P(X>3) =0.05.

10025 = 2,500 periods to consider, giving 2500 P(X>3) = 125 lost bonuses.

Question 17

What type of values can a continuous random variable assume?

Answer 17

Any value x in an interval on the real line or in a collection of intervals.

Question 18

Can we find the probability that a continuous random variable equals a specific value?

Answer 18

No, the probability that it equals a specific value is 0.

Question 19

How do we express probabilities for a continuous random variable?

Answer 19

We talk about the probability that it falls within a given interval.

Question 20

How is the probability defined for a continuous random variable between two values x_1 and x_2?

Answer 20

It is the area under the probability density function (PDF) between x_1 and x_2

Question 21

What is a probability density function (PDF)?

Answer 21

A function whose area under the curve between two points represents the probability of the variable falling within that interval.

Question 22

What is a uniform distribution in the context of continuous variables?

Answer 22

A distribution where all intervals of the same length are equally likely; the PDF is flat

Question 23

What is a normal distribution?

Answer 23

A bell-shaped, symmetric distribution defined by its mean (μ) and standard deviation (σ), with most values near the mean.

Question 24

In a normal distribution, what does the area under the curve represent?

Answer 24

The probability that the variable falls within a specific interval.

Question 25

When is a random variable said to be uniformly distributed?

Answer 25

When the probability is proportional to the length of the interval.

Question 26

What is the uniform probability density function (PDF)?

Answer 26

f(x) = 1/(b-a) = 0 for a ≤ X ≤ b elsewhere where: a = smallest value the variable can assume b = largest value the variable can assume

Question 27

What does the graph of a uniform distribution look like?

Answer 27

A horizontal line between a and b, representing equal probability across the interval.

Question 28

What is the expected value (mean) of a uniform distribution?

Answer 28

E(X) = (a+b)/2 (The average of the smallest and largest values.)

Question 29

What is the variance of a uniform distribution?

Answer 29

Var(X) = (b-a)ˆ2 /12

Question 30

In a uniform distribution, what do a and be represent again?

Answer 30

a: minimum possible value b: maximum possible value

Question 31

Example Stein customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 140g and 420g. Uniform Probability Density Function f(x) = 1/280 for 140 < X < 420 = 0 elsewhere where: X = salad plate filling weight

Answer 31

Expected Value of X E(X) = (a + b)/2 = (140 + 420)/2 = 280 Variance of X Var(X) = (b - a)2/12 = (420 – 140)2/12 = 6533.33

Question 32

What is the probability that a customer will take between 336 and 420 grams of salad?

Answer 32

P(336 < X < 420) = 1/280(84) = 0.3

Question 33

Why is the normal distribution important?

Answer 33

It’s the most important distribution for describing continuous random variables and is widely used in statistical inference.

Question 34

What is the general shape of the normal distribution?

Answer 34

A bell-shaped, symmetric curve centered around the mean.

Question 35

What is the formula for the normal probability density function (PDF)?

Answer 35

f(x) = (1/ σ√2π ) (e)ˆ-(x-μ)ˆ2/ 2σˆ2 where: μ = mean; σ = standard deviation π = 3.14159 e = 2.71828

Question 36

What is the skewness of the normal distribution?

Answer 36

The skewness is zero, meaning the distribution is perfectly symmetric.

Question 37

Where is the highest point on the normal curve?

Answer 37

At the mean, which is also the median and the mode.

Question 38

In a normal distribution, how are the mean, median, and mode related?

Answer 38

They are all equal and located at the center of the distribution.

Question 39

What defines a specific normal distribution within the family of normal distributions?

Answer 39

Its mean (μ) and standard deviation (σ).

Question 40

Can the mean be any numerical value?

Answer 40

Yes, the mean can be negative, zero, or positive depending on the data values.

Question 41

What does the standard deviation determine in a normal distribution?

Answer 41

The standard deviation determines the width of the curve. Larger values result in wider, flatter curves, indicating more variability in the data. Smaller values result in narrower, taller curves, indicating less variability.

Question 42

How are probabilities represented for a normal random variable?

Answer 42

Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1, with 0.5 to the left of the mean and 0.5 to the right.

Question 43

What percentage of values fall within certain standard deviations of the mean in a normal distribution?

Answer 43

68.26% of values are within +/- 1 standard deviation of the mean. 95.44% of values are within +/- 2 standard deviations of the mean. 99.72% of values are within +/- 3 standard deviations of the mean.

Question 44

What is a standard normal probability distribution?

Answer 44

A random variable with a normal distribution having a mean of 0 and a standard deviation of 1 is called a standard normal probability distribution. The letter Z is used to designate the standard normal random variable.

Question 45

What does the Z value represent in a standard normal distribution?

Answer 45

The Z value represents the number of standard deviations that a value (X) is from the mean. It is used to convert any normal distribution to the standard normal distribution. Z = (X - μ) / σ

Question 46

What is the standard normal density function?

Answer 46

f(x) = (1/ √2π ) (e)ˆ-(z)ˆ2 / 2 where: z = (x – m)/s π = 3.14159 e = 2.71828

Question 47

Example The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that demand during replenishment lead-time is normally distributed with a mean of 60 litres and a standard deviation of 24 litres. The manager would like to know the probability of a stockout, P(X > 80 litres).

Answer 47

Solving for the Stockout Probability: Step 1: Convert x to the standard normal distribution z = (x - μ)/ σ = (80 - 60)/24 = 0.83 Step 2: Find the area under the standard normal curve to the left of z = 0.83. Normal CD Lower = -10000 Upper = 0.83 σ = 1 μ = 0 P(Z≤0.83) = 0.796730 area P(Z>0.83) = 1- P(Z≤0.83) = 1 - 0.7967 = 0.2033 area Probability of a stockout, P(X>80)

Question 48

Example If the manager of Pep Zone wants the probability of a stockout to be no more than 0.05, what should the reorder point be?

Answer 48

Area = 0.95 left side Area = 0.05 right side Step 1: Find the z-value that cuts off an area of 0.05 in the right tail of the standard normal distribution. Normal CD Lower = -10000 Upper = 1.6 (trial and error) σ = 1 μ = 0 P(Z≤1.64) = 0.9495 area P(Z≤1.65) = 0.9505 area Step 2: Convert z_0.05 to the corresponding value of x. x = μ + z_0.05σ = 60 + 1.645(24) = 99.48 or 100 A reorder point of 100 litres will place the probability of a stockout during leadtime at (slightly less than) 0.05. By raising the reorder point from 80 litres to 100 litres on hand, the probability of a stockout decreases from about 0.20 to 0.05.

Question 49

When can the normal distribution be used to approximate binomial probabilities?

Answer 49

The normal distribution can approximate binomial probabilities when: n>20 nπ ≥ 5 n(1-π) ≥ 5 This is useful because calculating binomial probabilities directly becomes difficult for large n

Question 50

How do you apply the normal approximation to a binomial distribution?

Answer 50

Set μ = nπ and σ = √(𝑛π(1−π )) Use a continuity correction by adding and subtracting 0.5 since a continuous distribution is approximating a discrete one. For example, P(x = 40) is approximated by P(39.5 ≤ X ≤ 40.5).

Question 51

Example Shares in the investment bank Black Plainly (BP) are generally expected to mirror the performance of the market index. BP’s annualised volatility is estimated to be 10%. Suppose you believe that next year the return on the stock market as a whole will be 19.6%. Assuming that BP’s returns are normally distributed, how likely is it that Black Plainly shares will not increase in value?

Answer 51

P(x<= 0) = ? z = (0-19.6%)/10% = -1.96 P(z<=-1.96) = 2.5%

Question 52

The price of a security is approximately normally distributed. In the last year on about 20% of working days the price was less than or equal to 20. On 25% of working days the price was above 75. Find the mean and standard deviation of the price.

Answer 52

P(X<= 20) = 0.2 P(Z<= (20-µ)/σ) = 0.2 or (20-µ)/σ) = -0.84 --- (1) P(X>75) = 0.25 P(X<=75) = 1-.25 = 0.75 P(Z<= (75-µ)/σ)) = 0.75 Or, (75-µ)/σ = 0.68 --- (2) Solve for µ and σ: 20- µ = -0.84 σ 75- µ = 0.68 σ Therefore, σ = 36.2 and µ = 50.4

Question 53

What is VaR?

Answer 53

Estimate of the loss from a given position over a fixed time period that will be equaled or exceeded with a given probability

Question 54

What two equivalent interpretations does VaR have?

Answer 54

Worst Case Loss: over one day, there is a 95% probability that we will not lose more than $ yy An unlikely event: on average, in one out of every 20 days, we should expect to incur a loss greater than or equal to a certain amount

Question 55

What are the 3 key aspects of VaR?

Answer 55

VaR measures the minimum potential loss at the stated probability; the actual loss that could be incurred could be higher. VaR is associated with a stated degree of probability. Lowering the probability (increasing the confidence interval) increases the VaR. VaR measure is associated with a specific time period. Increasing the time interval will increase the VaR.

Question 56

Example We have a position worth $10 million in Microsoft shares The volatility of Microsoft is 2% per day The standard deviation of the change in the portfolio in 1 day is $200,000 Assuming the expected 1-day return is zero, what is the VaR at 99%?

Answer 56

P(x<=T) = 0.01 P(Z<= (T-0)/200,000)) = 0.01 ((T-0)/200,000) = -2.33 Or, T = -2.33*200,000 = -466,000 99% of the time, the portfolio’s loss will be no more than $466,000!