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Expected value and variability · Tutorial 334 of 1000

Mean of a Random Variable with a Missing Probability

Complete a discrete probability distribution before calculating its expected value and standard deviation.

Intermediate 9 min read

What You'll Learn

  • Find one missing probability by using the fact that a valid distribution totals 1.
  • Check that the completed probabilities are valid before calculating summaries.
  • Calculate a random variable’s mean from all values and their probabilities.
  • Find standard deviation using weighted squared deviations or a computational formula.
  • Verify calculations and interpret mean and standard deviation in context.
  • Recognize when the given information is not enough to determine missing probabilities.

Complete the Distribution Before Summarizing It

In Expected Value Versus a Single Outcome, you saw how the mean summarizes a distribution without predicting one particular result. To calculate that mean—or the standard deviation describing spread—you need a complete probability distribution. Sometimes a table gives every possible value but leaves one probability blank.

When exactly one probability is missing, the probabilities of all possible values must still add to 1. As explained in Finding a Missing Probability in a Distribution, subtract the sum of the known probabilities from 1 to find the missing entry. Then check the completed distribution before using it to find \(\mu_X\) and \(\sigma_X\).

Key idea: For a valid discrete probability distribution with exactly one missing probability, first use the total probability of 1 to fill the gap. Then use every value and its completed probability to calculate the mean and standard deviation.

The order matters. If you calculate a mean while treating the blank as 0, you have changed the model: the listed probabilities no longer account for all possible outcomes. A missing entry is unknown, not automatically zero. Also, a single total-probability equation determines one missing probability, but it generally cannot determine two or more unknown probabilities without additional information.

A Reliable Calculation Sequence

Let \(X\) be a discrete random variable with possible values \(x\). Once the distribution is complete, its mean is \(\mu_X=\sum xP(X=x)\). Its variance is \(\sigma_X^2=\sum (x-\mu_X)^2P(X=x)\), and its standard deviation is the nonnegative square root \(\sigma_X=\sqrt{\sigma_X^2}\). These are the methods from The Mean of a Discrete Random Variable, Variance of a Discrete Random Variable, and The Standard Deviation of a Discrete Random Variable.

1
Identify the outcomes and the blank.
Define \(X\), list every possible value, and mark the one probability that is unknown.
2
Complete the probability.
Add the known probabilities and subtract their sum from 1.
3
Check the distribution.
Confirm every probability is between 0 and 1, inclusive, and that all probabilities total 1. Check that the listed outcomes cover the model.
4
Calculate the mean and spread.
Use all value-probability pairs to find \(\mu_X\), then find \(\sigma_X^2\) and take its square root.
5
Interpret with units.
Describe \(\mu_X\) as the probability-weighted average and \(\sigma_X\) as the typical distance from that mean, in the original units.

A useful check for variance is the computational formula \(\sigma_X^2=\sum x^2P(X=x)-\mu_X^2\). It gives the same result as adding the weighted squared deviations, but can be quicker. Use it as a second calculation to catch arithmetic errors. Variance is in squared units; standard deviation is in the original units.

Worked Example: Refill Amounts with One Missing Probability

Worked Example: Refill Amounts with One Missing Probability

A fictional water station models the amount \(X\), in liters, dispensed in one refill. The possible amounts and probabilities are shown below, with one probability missing. Find the missing probability, mean, and standard deviation.

\(x\), liters\(P(X=x)\)
00.25
2?
50.35

State. Let \(X\) be the amount of water, in liters, dispensed during one refill. We need to complete its distribution and calculate \(\mu_X\) and \(\sigma_X\).

Plan. The three listed amounts represent all outcomes in this model. There is exactly one unknown probability, so use the requirement that a valid distribution totals 1. Then check the completed probabilities, calculate the weighted mean and squared deviations, and verify the variance with the computational formula.

Do. The missing probability is:

$$ P(X=2)=1-(0.25+0.35)=1-0.60=0.40. $$

The check is \(0.25+0.40+0.35=1.00\), and each probability is between 0 and 1. The distribution is valid. Now calculate the mean:

$$ \begin{aligned} \mu_X &=(0)(0.25)+(2)(0.40)+(5)(0.35)\\ &=0+0.80+1.75\\ &=2.55\text{ liters}. \end{aligned} $$

As a calculator check, entering the values and probabilities in 1-Var Stats, using the probabilities as the frequency list, gives \(\mu_X=2.55\). The weighted squared deviations give:

$$ \begin{aligned} \sigma_X^2 &=(0-2.55)^2(0.25)+(2-2.55)^2(0.40)+(5-2.55)^2(0.35)\\ &=(6.5025)(0.25)+(0.3025)(0.40)+(6.0025)(0.35)\\ &=1.625625+0.121+2.100875\\ &=3.8475\text{ liters}^2,\\ \sigma_X&=\sqrt{3.8475}\approx1.9615\text{ liters}. \end{aligned} $$

Check the variance with \(\sum x^2P(X=x)-\mu_X^2\):

$$ \begin{aligned} \sigma_X^2 &=(0^2)(0.25)+(2^2)(0.40)+(5^2)(0.35)-(2.55)^2\\ &=0+1.60+8.75-6.5025\\ &=3.8475\text{ liters}^2. \end{aligned} $$

The standard deviation check is \(\sqrt{3.8475}\approx1.9615\), matching the first calculation. 1-Var Stats with the same inputs also reports \(\sigma_X\approx1.9615\).

Conclude. Under this model, the average amount dispensed per refill is 2.55 liters. The standard deviation is about 1.96 liters, so refill amounts typically differ from their mean by about 1.96 liters. This describes the distribution’s center and spread, not a guarantee about any one refill.

Worked Example: Damaged Items in a Package

Worked Example: Damaged Items in a Package

A fictional quality-control model describes \(X\), the number of damaged items in a package of inspected goods. The package can have 0, 1, or 2 damaged items. The probability of 0 damaged items is 0.20, the probability of 2 is 0.30, and the probability of 1 is unknown. Find the missing probability, mean, and standard deviation.

The missing probability is \(1-(0.20+0.30)=0.50\). The completed probabilities are all between 0 and 1 and sum to \(0.20+0.50+0.30=1.00\). The three possible counts cover this model.

Calculate the mean:

$$ \begin{aligned} \mu_X &=(0)(0.20)+(1)(0.50)+(2)(0.30)\\ &=0+0.50+0.60\\ &=1.10\text{ damaged items}. \end{aligned} $$

For a second check, 1-Var Stats with the values and probabilities as frequencies gives a mean of 1.10. For the variance:

$$ \begin{aligned} \sigma_X^2 &=(0-1.10)^2(0.20)+(1-1.10)^2(0.50)+(2-1.10)^2(0.30)\\ &=(1.21)(0.20)+(0.01)(0.50)+(0.81)(0.30)\\ &=0.242+0.005+0.243\\ &=0.49\text{ items}^2. \end{aligned} $$

Verify with the computational formula:

$$ \begin{aligned} \sigma_X^2 &=(0^2)(0.20)+(1^2)(0.50)+(2^2)(0.30)-(1.10)^2\\ &=0+0.50+1.20-1.21\\ &=0.49\text{ items}^2,\\ \sigma_X&=\sqrt{0.49}=0.70\text{ damaged items}. \end{aligned} $$

Both methods give a standard deviation of 0.70 items; 1-Var Stats confirms \(\sigma_X=0.70\). Under this model, the average number of damaged items per package is 1.10, and counts typically differ from that average by about 0.70 item. The mean need not be a possible outcome: a package cannot contain 1.10 damaged items, but the probability-weighted average can be 1.10.

Worked Example: A Missing Probability in a Waiting-Time Model

Worked Example: A Missing Probability in a Waiting-Time Model

A fictional shuttle service uses a discrete model for \(X\), a passenger’s waiting time, in minutes, at a particular stop. The modeled times are 1, 3, and 7 minutes, with probabilities 0.20, an unknown value, and 0.30, respectively. Find the missing probability, mean, and standard deviation.

The blank is \(1-(0.20+0.30)=0.50\). The completed distribution has probabilities \(0.20\), \(0.50\), and \(0.30\), which are each valid and total 1.00. All three stated waiting times are included in the model.

The mean waiting time is:

$$ \begin{aligned} \mu_X &=(1)(0.20)+(3)(0.50)+(7)(0.30)\\ &=0.20+1.50+2.10\\ &=3.80\text{ minutes}. \end{aligned} $$

As a check, 1-Var Stats using the waiting times as values and probabilities as frequencies gives \(\mu_X=3.80\). The variance from squared deviations is:

$$ \begin{aligned} \sigma_X^2 &=(1-3.80)^2(0.20)+(3-3.80)^2(0.50)+(7-3.80)^2(0.30)\\ &=(7.84)(0.20)+(0.64)(0.50)+(10.24)(0.30)\\ &=1.568+0.320+3.072\\ &=4.96\text{ minutes}^2. \end{aligned} $$

Use the computational formula as a second variance calculation:

$$ \begin{aligned} \sigma_X^2 &=(1^2)(0.20)+(3^2)(0.50)+(7^2)(0.30)-(3.80)^2\\ &=0.20+4.50+14.70-14.44\\ &=4.96\text{ minutes}^2,\\ \sigma_X&=\sqrt{4.96}\approx2.2271\text{ minutes}. \end{aligned} $$

The square-root check gives \(\sqrt{4.96}\approx2.2271\), and 1-Var Stats confirms \(\sigma_X\approx2.2271\). The modeled average wait is 3.80 minutes. Waiting times typically differ from that mean by about 2.23 minutes. This mean is not necessarily a waiting time an individual passenger experiences.

Common Mistakes and AP Exam Tips

  • Using the blank as zero. An omitted probability is unknown until you calculate it. A full-credit response shows \(1-\text{(sum of known probabilities)}\) and gives the resulting probability.
  • Finding the mean before completing the distribution. Calculate \(\mu_X\) only after the probability table is valid. Otherwise, the weighted average does not represent the stated model.
  • Assuming the missing probability is always possible. Check that the result is between 0 and 1. If it is negative or greater than 1, the stated information is inconsistent with a valid distribution.
  • Assuming total probability is enough for multiple blanks. One equation stating that probabilities total 1 cannot usually determine two unknowns. Explain that more information is needed instead of choosing values without a basis.
  • Reporting variance as standard deviation. Variance is in squared units. Take its square root to report standard deviation in the original units.
  • Forgetting context and units. State what \(X\) measures. Interpret \(\mu_X\) as the probability-weighted average and \(\sigma_X\) as a typical distance from that mean—not as a guarantee for one outcome.

For a complete response, show the missing-probability calculation, verify that the finished distribution is valid, and include the products used to calculate the mean. Then show a variance method and the square root. A second variance calculation using \(\sum x^2P(X=x)-\mu_X^2\) is a useful arithmetic check.

Key Takeaway

A missing probability must be resolved before the distribution can support a mean or standard deviation. Once the distribution is complete and valid, use every value-probability pair, check the calculations, and interpret the results with the variable’s units.

Key takeaway: Find one missing probability by making the total 1. Then calculate \(\mu_X=\sum xP(X=x)\) and \(\sigma_X=\sqrt{\sum (x-\mu_X)^2P(X=x)}\). Check the distribution and keep variance, standard deviation, and their units distinct.

Check Your Understanding

A fictional device records \(X\), the number of alerts during one hour. Its possible values are 0, 1, and 3, with probabilities 0.35, an unknown value, and 0.25.

  1. Find the missing probability and show how you checked that the completed probabilities total 1.
  2. Calculate \(\mu_X\), the mean number of alerts per hour.
  3. Calculate \(\sigma_X^2\) using weighted squared deviations, then find \(\sigma_X\).
  4. Verify the variance with \(\sum x^2P(X=x)-\mu_X^2\).
  5. Interpret the mean and standard deviation in context. Explain why the mean does not have to be a possible number of alerts in one hour.

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