Math Problem Statement

Cookies have weights that are normally distributed with a mean of 75 grams and a standard deviation of 8 grams. They are sold in packages containing 13 randomly selected cookies.

a) What percentage of the cookies weighs more than 68 grams?

b) What is the probability that the total weight of the cookies in a package is less than 900 grams?

Enter your answers below rounded to three decimal places if necessary.

a) Answer 1 Question 31 b) Answer 2 Question 31

Solution

Let’s solve each part step by step:


a) Percentage of cookies weighing more than 68 grams

The weights of the cookies follow a normal distribution with:

μ=75grams,σ=8grams\mu = 75 \, \text{grams}, \, \sigma = 8 \, \text{grams}

We need to find P(X>68)P(X > 68), where XX is the weight of a single cookie.

  1. Standardize XX: Z=XμσZ = \frac{X - \mu}{\sigma} Substituting X=68X = 68: Z=68758=78=0.875Z = \frac{68 - 75}{8} = \frac{-7}{8} = -0.875

  2. Find the probability: Using a standard normal table or calculator, find P(Z>0.875)P(Z > -0.875). For symmetry, we use: P(Z>0.875)=1P(Z0.875)P(Z > -0.875) = 1 - P(Z \leq -0.875)

    From standard normal distribution tables: P(Z0.875)0.1908P(Z \leq -0.875) \approx 0.1908

    So: P(Z>0.875)=10.1908=0.8092P(Z > -0.875) = 1 - 0.1908 = 0.8092

    Therefore, 80.92% of the cookies weigh more than 68 grams.


b) Probability that the total weight of cookies in a package is less than 900 grams

The total weight of 13 cookies is the sum of 13 independent normally distributed random variables. The sum also follows a normal distribution with:

Mean of total weight: μtotal=nμ=1375=975grams\text{Mean of total weight: } \mu_{\text{total}} = n \cdot \mu = 13 \cdot 75 = 975 \, \text{grams} Standard deviation of total weight: σtotal=nσ=13828.844grams\text{Standard deviation of total weight: } \sigma_{\text{total}} = \sqrt{n} \cdot \sigma = \sqrt{13} \cdot 8 \approx 28.844 \, \text{grams}

We need P(Total Weight<900)P(\text{Total Weight} < 900), or P(Y<900)P(Y < 900), where YY is the total weight.

  1. Standardize YY: Z=YμtotalσtotalZ = \frac{Y - \mu_{\text{total}}}{\sigma_{\text{total}}} Substituting Y=900Y = 900: Z=90097528.844=7528.8442.600Z = \frac{900 - 975}{28.844} = \frac{-75}{28.844} \approx -2.600

  2. Find the probability: From standard normal distribution tables: P(Z<2.600)0.0047P(Z < -2.600) \approx 0.0047

    Therefore, the probability is approximately 0.0047 (0.47%).


Final Answers:

a) 80.920
b) 0.005

Let me know if you want further clarifications or a step-by-step explanation of any part!


Related Questions:

  1. What is the probability that a single cookie weighs less than 60 grams?
  2. How does increasing the sample size affect the standard deviation of the total weight?
  3. What percentage of the cookies weigh between 70 and 80 grams?
  4. What is the z-score for a cookie weighing 85 grams?
  5. What assumptions are made about the distribution of cookie weights?

Tip:

To quickly compute probabilities for normal distributions, tools like z-tables, calculators, or software like Excel can be extremely helpful!

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Standardization using Z-scores
Probability and Statistics

Formulas

Z = (X - μ) / σ
Probability from standard normal distribution tables
Mean of total weight: μ_total = n × μ
Standard deviation of total weight: σ_total = √n × σ

Theorems

Central Limit Theorem

Suitable Grade Level

Grades 10-12