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

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:

\[
\mu = 75 \, \text{grams}, \, \sigma = 8 \, \text{grams}
\]

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

1. **Standardize \( X \):**
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   Substituting \( X = 68 \):
   \[
   Z = \frac{68 - 75}{8} = \frac{-7}{8} = -0.875
   \]

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

   From standard normal distribution tables:
   \[
   P(Z \leq -0.875) \approx 0.1908
   \]

   So:
   \[
   P(Z > -0.875) = 1 - 0.1908 = 0.8092
   \]

   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:

\[
\text{Mean of total weight: } \mu_{\text{total}} = n \cdot \mu = 13 \cdot 75 = 975 \, \text{grams}
\]
\[
\text{Standard deviation of total weight: } \sigma_{\text{total}} = \sqrt{n} \cdot \sigma = \sqrt{13} \cdot 8 \approx 28.844 \, \text{grams}
\]

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

1. **Standardize \( Y \):**
   \[
   Z = \frac{Y - \mu_{\text{total}}}{\sigma_{\text{total}}}
   \]
   Substituting \( Y = 900 \):
   \[
   Z = \frac{900 - 975}{28.844} = \frac{-75}{28.844} \approx -2.600
   \]

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

   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!

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

Explore the normal distribution with this problem involving cookie weights. Learn to calculate the percentage of cookies weighing more than 68 grams and the probability that the total weight of 13 cookies is less than 900 grams. This detailed solution includes Z-scores, standardization, and step-by-step probability computations. Suitable for Grades 10-12.

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