Math Problem Statement

Assume that a procedure yields a binomial distribution with n=1301 trials and the probability of success for one trial is p=72%.

Find the mean for this binomial distribution. (Round answer to one decimal place.) μ= 

Find the standard deviation for this distribution. (Round answer to two decimal places.) σ= 

Use the range rule of thumb to find the minimum usual value μ–2σ and the maximum usual value μ+2σ. Enter answer as an interval using square-brackets only with whole numbers. usual values = 

Solution

To solve this problem, we will use the formulas for the mean and standard deviation of a binomial distribution. The mean μ\mu and standard deviation σ\sigma for a binomial distribution are given by:

  1. Mean μ=n×p\mu = n \times p
  2. Standard deviation σ=n×p×(1p)\sigma = \sqrt{n \times p \times (1 - p)}

Where:

  • n=1301n = 1301 (the number of trials)
  • p=0.72p = 0.72 (the probability of success)

Step 1: Find the Mean (μ\mu)

μ=n×p=1301×0.72=936.72\mu = n \times p = 1301 \times 0.72 = 936.72

So, the mean μ\mu is approximately:

μ=936.7\mu = 936.7

Step 2: Find the Standard Deviation (σ\sigma)

σ=n×p×(1p)=1301×0.72×(10.72)=1301×0.72×0.28\sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{1301 \times 0.72 \times (1 - 0.72)} = \sqrt{1301 \times 0.72 \times 0.28}

σ=1301×0.2016=262.1716.2\sigma = \sqrt{1301 \times 0.2016} = \sqrt{262.17} \approx 16.2

So, the standard deviation σ\sigma is approximately:

σ=16.2\sigma = 16.2

Step 3: Find the Usual Range Using the Range Rule of Thumb

The usual values are calculated using:

μ2σandμ+2σ\mu - 2\sigma \quad \text{and} \quad \mu + 2\sigma

For the minimum usual value:

μ2σ=936.72×16.2=936.732.4=904.3\mu - 2\sigma = 936.7 - 2 \times 16.2 = 936.7 - 32.4 = 904.3

For the maximum usual value:

μ+2σ=936.7+2×16.2=936.7+32.4=969.1\mu + 2\sigma = 936.7 + 2 \times 16.2 = 936.7 + 32.4 = 969.1

Thus, the usual values fall within the interval:

usual values=[904,969]\text{usual values} = [904, 969]

Final Answers:

  • Mean (μ\mu): 936.7
  • Standard deviation (σ\sigma): 16.2
  • Usual values interval: [904, 969]

Let me know if you'd like more details or have any questions! Also, here are some related questions:

  1. What is the variance of this binomial distribution?
  2. How would you calculate the probability of getting exactly 900 successes in this binomial distribution?
  3. How can the binomial distribution be approximated by a normal distribution in this case?
  4. What is the probability of getting fewer than 800 successes?
  5. How does increasing or decreasing the number of trials nn affect the standard deviation of a binomial distribution?

Tip: The mean of a binomial distribution is helpful for determining the expected number of successes in a given number of trials.

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Math Problem Analysis

Mathematical Concepts

Binomial Distribution
Probability
Standard Deviation
Mean

Formulas

Mean: μ = n × p
Standard Deviation: σ = √(n × p × (1 - p))
Usual Values: μ - 2σ, μ + 2σ

Theorems

Range Rule of Thumb for Usual Values

Suitable Grade Level

Grades 11-12