Math Problem Statement

The probabilities that a patient will have 0, 1, 2, or 3 medical tests performed on entering a hospital are ! "# , # "# , $ "# , and " "# , respectively

Solution

The probabilities provided for a patient having 0, 1, 2, or 3 medical tests upon entering a hospital are:

  • Probability of 0 tests: 18\frac{1}{8}
  • Probability of 1 test: 38\frac{3}{8}
  • Probability of 2 tests: 38\frac{3}{8}
  • Probability of 3 tests: 18\frac{1}{8}

These probabilities form a discrete probability distribution, which can be used to compute several important statistics such as the expected number of tests, variance, and standard deviation.

1. Expected Value (Mean)

The expected value E(X)E(X) of the number of tests XX can be calculated using the formula:

E(X)=xxP(x)E(X) = \sum_{x} x \cdot P(x)

Where xx is the number of tests, and P(x)P(x) is the probability associated with xx.

So, for this case:

E(X)=018+138+238+318E(X) = 0 \cdot \frac{1}{8} + 1 \cdot \frac{3}{8} + 2 \cdot \frac{3}{8} + 3 \cdot \frac{1}{8}

E(X)=0+38+68+38=128=1.5E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = 1.5

2. Variance

The variance Var(X)\text{Var}(X) is calculated using:

Var(X)=E(X2)[E(X)]2\text{Var}(X) = E(X^2) - [E(X)]^2

First, let's calculate E(X2)E(X^2):

E(X2)=0218+1238+2238+3218E(X^2) = 0^2 \cdot \frac{1}{8} + 1^2 \cdot \frac{3}{8} + 2^2 \cdot \frac{3}{8} + 3^2 \cdot \frac{1}{8}

E(X2)=0+38+128+98=248=3E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3

Now, the variance is:

Var(X)=E(X2)[E(X)]2=3(1.5)2=32.25=0.75\text{Var}(X) = E(X^2) - [E(X)]^2 = 3 - (1.5)^2 = 3 - 2.25 = 0.75

3. Standard Deviation

The standard deviation σ\sigma is the square root of the variance:

σ=Var(X)=0.750.866\sigma = \sqrt{\text{Var}(X)} = \sqrt{0.75} \approx 0.866

Would you like more details on any of these calculations?

Here are some related questions:

  1. How does the expected number of tests change with different probability distributions?
  2. What is the importance of variance in this context?
  3. How would the calculations change if the probabilities were not equally distributed?
  4. How can we interpret the standard deviation in a medical context?
  5. How would you calculate the probability of having more than 1 test performed?

Tip: The expected value tells you the average outcome in the long run if the scenario (in this case, testing) is repeated many times.

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Math Problem Analysis

Mathematical Concepts

Probability
Statistics
Discrete Probability Distribution

Formulas

Expected value: E(X) = ∑x * P(x)
Variance: Var(X) = E(X^2) - [E(X)]^2
Standard deviation: σ = √Var(X)

Theorems

Law of Total Expectation
Variance Theorem

Suitable Grade Level

Grades 10-12