Interval, conditional and μ ± 2σ probabilities from a discrete distribution

What should be done first for an event probability from a discrete distribution?

Move the upper boundary and compare values at most the boundary with values strictly below it. A row belongs to the first event when its final column is zero or negative; the second event requires a negative entry. Add only the probabilities of included rows. The displayed probabilities are decimals for reading; the exact masses are one tenth, one fifth, three tenths, one quarter and three twentieths

At t=2, how much larger is Pr(X≤t) than Pr(X<t)?

Event probabilities for a discrete random variable

For a discrete event, select the values in the table that satisfy the stated condition and add their probabilities. Conditional probability uses the intersection in the numerator and the conditioning event in the denominator

1. 1Read the event exactly Include a non-strict endpoint and exclude a strict endpoint \(\displaystyle x\in\{1,2,3,5,6\},\quad2<x\leq5\Rightarrow\{3,5\}\)
2. 2Add selected probabilities The possible values in one event are disjoint, so add their probabilities \(\displaystyle \Pr(X\in\{3,5\})=\frac{1}{4}+\frac{3}{8}=\frac{5}{8}\)
3. 3Use the conditioning event For Pr(A given B), divide by the probability of B. \(\displaystyle \Pr(A\mid B)=\frac{\Pr(A\cap B)}{\Pr(B)}\)
4. 4Translate two standard deviations Calculate μ - 2σ and μ + 2σ, then keep only table values between the bounds \(\displaystyle \mu=4,\quad\sigma=1\Rightarrow\mu-2\sigma=2,\quad\mu+2\sigma=6\)

Complete distribution

\(\displaystyle \begin{array}{c|cccc}x&0&1&3&5\\\Pr(X=x)&1/8&1/4&3/8&1/4\end{array}\)

For the event 1 < X <= 5 on values 0, 1, 3 and 5, read the probability for X = 3.

Complete distribution

\(\displaystyle \begin{array}{c|cccc}x&0&1&3&5\\\Pr(X=x)&1/8&1/4&3/8&1/4\end{array}\)

Add the probabilities for X = 3 and X = 5 to find Pr(1 < X <= 5)

Multiplying probabilities for different values. One of these lines is right. Which one?

Select values before adding their probabilities

For this score model, consider −1<X≤5. The lower endpoint −1 is excluded and the upper endpoint 5 is included. The qualifying values are 2 and 5; no other real number between the bounds is a possible value of X.

Distribution and requested quantity

\(\displaystyle \begin{array}{c|ccccc}x&-4&-1&2&5&8\\\Pr(X=x)&1/10&1/5&3/10&1/4&3/20\end{array}\)

Calculate Pr(−1<X≤5) from this distribution

Distribution and requested quantity

\(\displaystyle \begin{array}{c|ccccc}x&-4&-1&2&5&8\\\Pr(X=x)&1/10&1/5&3/10&1/4&3/20\end{array}\)

Change only the upper boundary to strict. Calculate Pr(−1<X<5)

Distribution and requested quantity

\(\displaystyle \begin{array}{c|ccccc}x&-4&-1&2&5&8\\\Pr(X=x)&1/10&1/5&3/10&1/4&3/20\end{array}\)

Find the probability of the complement of −1<X≤5 within this distribution

Restrict to the conditioning event

Use the same distribution. Let A be X≥2 and B be X≤5. List A, B and their intersection before calculating. The probability Pr(A given B) considers only the outcomes in B, so the denominator must be Pr(B)>0.

Distribution and requested quantity

\(\displaystyle \begin{array}{c|ccccc}x&-4&-1&2&5&8\\\Pr(X=x)&1/10&1/5&3/10&1/4&3/20\end{array}\)

Calculate Pr(A∩B), where A is X≥2 and B is X≤5.

Distribution and requested quantity

\(\displaystyle \begin{array}{c|ccccc}x&-4&-1&2&5&8\\\Pr(X=x)&1/10&1/5&3/10&1/4&3/20\end{array}\)

Calculate Pr(B) for B given by X≤5.

Intersection probability

\(\displaystyle \Pr(A\cap B)=\frac{11}{20}\)

Conditioning probability

\(\displaystyle \Pr(B)=\frac{17}{20}\)

Calculate Pr(A given B)

Reversing the conditioning changes the denominator

Let A be X≥2 and B be X≤5. For Pr(B given A), the intersection is unchanged but the conditioning event is now A. Its values 2,5,8 have total probability 7/10. There is no general rule that Pr(A given B)=Pr(B given A)

Intersection probability

\(\displaystyle \Pr(A\cap B)=\frac{11}{20}\)

Conditioning probability

\(\displaystyle \Pr(A)=\frac7{10}\)

Calculate Pr(B given A)

Check the conditional distribution

Inside B, divide each original mass by the probability of B. The conditional probabilities in the table sum to one. Adding the last two gives the probability of A given B.

Possible values of X

\(\displaystyle X\in\{-4,-1,2,5,8\}\)

Let A be X≥2 and C be X>8. Can the formula define Pr(A given C)?

Calculate a two-standard-deviation event from the masses

A different distribution has values −9,−3,3,9,15 with probabilities 1/36,1/9,13/18,1/9,1/36. Calculate its mean, variance and standard deviation, then select values inside the inclusive interval μ−2σ≤X≤μ+2σ

Distribution and requested quantity

\(\displaystyle (-9)\frac1{36}+(-3)\frac19+3\frac{13}{18}+9\frac19+15\frac1{36}\)

Calculate the mean of this second distribution

Distribution and requested quantity

\(\displaystyle \mu=3,\quad\operatorname{E}(X^2)=25\)

Find the variance using the exact mean and expected square

Distribution and requested quantity

\(\displaystyle \operatorname{Var}(X)=16\)

Find the non-negative standard deviation

Distribution and requested quantity

\(\displaystyle \mu=3,\quad\sigma=4\)

Calculate the exact lower bound μ−2σ

Distribution and requested quantity

\(\displaystyle \mu=3,\quad\sigma=4\)

Calculate the exact upper bound μ+2σ

Distribution

\(\displaystyle \begin{array}{c|ccccc}x&-9&-3&3&9&15\\\Pr(X=x)&1/36&1/9&13/18&1/9&1/36\end{array}\)

Interval

\(\displaystyle [-5,11]\)

Find Pr(μ−2σ≤X≤μ+2σ) for the second distribution

Do not substitute a normal-model percentage

This discrete interval has probability 17/18, not an assumed 0.95. The one-standard-deviation interval [−1,7] contains only 3 and has probability 13/18. Extending the two-standard-deviation bounds outward to −9 and 15 would incorrectly include every value. This event is an interval of values of X, not a confidence interval for an unknown parameter

Keep exact bounds when the standard deviation is irrational

Another model has values −7,−2,3,8,13 with probabilities 1/32,1/8,11/16,1/8,1/32. Its mean is 3 and variance is 25/2, so the two-standard-deviation bounds are 3−5√2 and 3+5√2. Give decimal displays only after deciding membership from the exact bounds

Exact mean and standard deviation

\(\displaystyle \mu=3,\quad\sigma=\frac{5\sqrt2}{2}\)

Decimal approximation

\(\displaystyle \sqrt2\approx1.41421356\)

Display μ−2σ correct to two decimal places

Exact mean and standard deviation

\(\displaystyle \mu=3,\quad\sigma=\frac{5\sqrt2}{2}\)

Decimal approximation

\(\displaystyle \sqrt2\approx1.41421356\)

Display μ+2σ correct to two decimal places

Verify membership without an early rounding decision

Since 5<5√2<10, the lower bound lies strictly between −7 and −2 and the upper bound lies strictly between 8 and 13. Only −2,3,8 are included, with total probability 15/16. If a possible value is close to a bound, compare with the exact bound rather than using its rounded display

Write a complete event and conditional analysis, ungraded

For values −6,0,3,9 with probabilities 1/8,3/8,1/4,1/4, let A be X>0 and B be X≤3. List A, B, their intersection and the complement of A. Calculate Pr(A given B) and Pr(B given A), construct the complete conditional distribution given B, and explain the effect of making the upper boundary X<3. This whole analysis is self-reviewed rather than graded by the scalar checks

Check the event sets and conditional totals

Let A be X>0 and B be X≤3. A={3,9}, B={−6,0,3}, A∩B={3} and the complement of A is {−6,0}. Pr(A)=1/2, Pr(B)=3/4 and Pr(A∩B)=1/4. Thus Pr(A given B)=1/3 and Pr(B given A)=1/2. Given B, the values −6,0,3 have probabilities 1/6,1/2,1/3. Replacing B by X<3 leaves {−6,0}; its intersection with A is empty and its positive probability is 1/2, so Pr(A given X<3)=0.

Check the event, the condition and the domain

Use only possible values from the distribution. Add mutually exclusive value masses, keep strict endpoints out, and divide by a positive conditioning probability when a condition is present. For μ±2σ, use exact bounds to select values and calculate their actual probability; rounding is a display step