Math Problem Statement

The probabilities that a patient will have 0,1 ,2 , or 3 medical tests performed on entering a hospital are 615 , 5/15 , 3/15 , and 1/15 respectively. Construct a probability distribution.

Medical Tests x P(x) (Fractions only) 0 Answer 1 Question 3 6/15 1 Answer 2 Question 3 5/15 2 Answer 3 Question 3 3/15 3 Answer 4 Question 3 1/15 What is the probability of 1 medical test being performed? Simplify all fractions.

P(x= Answer 5 Question 3 1 )= Answer 6 Question 3 1/3

What is the probability of less than 2 medical tests being performed?

P(x< Answer 7 Question 3 2 )= Answer 8 Question 3 11/15

What is the probability of between 0 and 1 tests being performed, inclusive?

P( Answer 9 Question 3 0 ≤x≤ Answer 10 Question 3 2 )= Answer 11 Question 3

What is the mean and standard deviation? Use decimals and round to the nearest thousandth.

μ= Answer 12 Question 3 0.933

σ= Answer 13 Question 3

Solution

Let's go step-by-step through the problem.

1. Probability Distribution

We are given the probabilities for 0, 1, 2, and 3 medical tests. Here's the probability distribution:

Medical Tests xxProbability P(x)P(x)
0615\frac{6}{15}
1515\frac{5}{15}
2315\frac{3}{15}
3115\frac{1}{15}

2. What is the probability of 1 medical test being performed?

The probability of exactly 1 test being performed is: P(x=1)=515=13P(x = 1) = \frac{5}{15} = \frac{1}{3}

3. What is the probability of less than 2 medical tests being performed?

To find P(x<2)P(x < 2), we sum the probabilities for 0 and 1 medical tests: P(x<2)=P(x=0)+P(x=1)=615+515=1115P(x < 2) = P(x = 0) + P(x = 1) = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}

4. What is the probability of between 0 and 1 tests being performed, inclusive?

To find P(0x1)P(0 \leq x \leq 1), sum the probabilities for 0 and 1 medical tests: P(0x1)=P(x=0)+P(x=1)=615+515=1115P(0 \leq x \leq 1) = P(x = 0) + P(x = 1) = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}

5. Mean ( μ\mu ) and Standard Deviation ( σ\sigma )

To find the mean and standard deviation, we will use the formulas:

  • Mean: μ=xP(x)\mu = \sum x P(x) Plugging in the values: μ=0×615+1×515+2×315+3×115=0+515+615+315=14150.933\mu = 0 \times \frac{6}{15} + 1 \times \frac{5}{15} + 2 \times \frac{3}{15} + 3 \times \frac{1}{15} = 0 + \frac{5}{15} + \frac{6}{15} + \frac{3}{15} = \frac{14}{15} \approx 0.933

  • Standard deviation: σ=(xμ)2P(x)\sigma = \sqrt{\sum (x - \mu)^2 P(x)} Using the mean μ=0.933\mu = 0.933, let's calculate σ\sigma.

I'll calculate the standard deviation next.The mean μ\mu is approximately 0.9330.933, and the standard deviation σ\sigma is approximately 0.9290.929, both rounded to the nearest thousandth.

Summary:

  1. Probability of 1 medical test: 13\frac{1}{3}
  2. Probability of less than 2 tests: 1115\frac{11}{15}
  3. Probability of 0 to 1 tests (inclusive): 1115\frac{11}{15}
  4. Mean ( μ\mu ): 0.9330.933
  5. Standard Deviation ( σ\sigma ): 0.9290.929

Would you like to dive deeper into any part of this? Here are some related questions:

  1. What is the cumulative probability distribution for each xx?
  2. How is the variance different from the standard deviation?
  3. How would this distribution change if one of the probabilities increased?
  4. What is the probability of getting more than 1 test?
  5. How can we visualize this probability distribution?

Tip: When working with probabilities, always ensure that the total probability adds up to 1!

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Math Problem Analysis

Mathematical Concepts

Probability Distribution
Mean (Expected Value)
Standard Deviation

Formulas

P(x) = P(x = 0) + P(x = 1) for cumulative probability
Mean: μ = ΣxP(x)
Standard Deviation: σ = √Σ((x - μ)^2 P(x))

Theorems

Law of Total Probability

Suitable Grade Level

Grades 10-12