Math Problem Statement

According to the South Dakota Department of Health, the mean number of hours of TV viewing per week is higher among adult women than men. A recent study showed women spent an average of 34 hours per week watching TV and men 29 hours per week (state.sd.us/DOH/Nutrition/TV.pdf). YOU DO NOT NEED TO GO TO THE LINK TO COMPLETE THE PROBLEM!

Assume that the distribution of hours watched follows the normal distribution for both groups, and that the standard deviation among the women is 4.5 hours and is 5.1 hours for the men.

What is the probability that a woman watches more than 30 hours of TV per week?

Please provide all six steps as shown in course materials. Upload your work.

Solution

To solve this problem, we'll calculate the probability that a woman watches more than 30 hours of TV per week, using the normal distribution and the given mean and standard deviation.

Six steps to solve the problem:

1. Identify the population of interest:

  • Population: Women who watch TV.
  • We're asked to find the probability that a randomly selected woman watches more than 30 hours of TV per week.

2. Identify the random variable of interest:

  • Let XX be the number of hours a woman watches TV per week.
  • XN(μ,σ2)X \sim N(\mu, \sigma^2), where μ=34\mu = 34 hours and σ=4.5\sigma = 4.5 hours.

3. State the probability statement:

  • We're tasked with finding P(X>30)P(X > 30), where XX follows the normal distribution with mean μ=34\mu = 34 and standard deviation σ=4.5\sigma = 4.5.

4. Convert to the standard normal distribution:

  • To calculate this probability, we'll convert the normal random variable XX into a standard normal variable ZZ, using the formula: Z=XμσZ = \frac{X - \mu}{\sigma} Substituting the values: Z=30344.5=44.50.889Z = \frac{30 - 34}{4.5} = \frac{-4}{4.5} \approx -0.889

5. Use the standard normal table:

  • Now, we'll find the probability P(Z>0.889)P(Z > -0.889).
  • First, look up the probability for Z=0.889Z = -0.889 in a standard normal distribution table or use a calculator. This gives: P(Z0.889)0.1879P(Z \leq -0.889) \approx 0.1879
  • To find P(Z>0.889)P(Z > -0.889), subtract this from 1: P(Z>0.889)=10.1879=0.8121P(Z > -0.889) = 1 - 0.1879 = 0.8121

6. State the conclusion:

  • The probability that a randomly selected woman watches more than 30 hours of TV per week is approximately 0.8121 or 81.21%.

Let me know if you'd like me to upload a step-by-step solution, including the work on paper or further clarifications.

Follow-up questions:

  1. What is the probability that a woman watches less than 30 hours of TV per week?
  2. How would the probability change if the average number of hours a woman watches TV increases to 36 hours?
  3. What is the probability that a man watches more than 30 hours of TV per week, using the same steps?
  4. If the standard deviation for women decreases to 3 hours, how does that affect the probability of watching more than 30 hours of TV?
  5. How would we calculate the probability for a woman watching between 30 and 40 hours of TV per week?

Tip:

For solving probability problems involving normal distributions, always remember to convert the raw score to a standard score ZZ, which simplifies the use of probability tables or software.

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Math Problem Analysis

Mathematical Concepts

Statistics
Probability Theory
Normal Distribution
Z-scores

Formulas

Z = (X - μ) / σ
P(Z > z-value) = 1 - P(Z ≤ z-value)

Theorems

Standard Normal Distribution Theorem

Suitable Grade Level

Grades 10-12