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The image provided contains a problem about the weight distribution of jars of jam. Here is the breakdown of the question and the relevant parts:

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**Problem Statement:**

Jam is sold in jars, and the mean weight of the contents is 108 grams. Only 3% of jars have contents weighing less than 100 grams. Assuming that the weight of jam in a jar is normally distributed, find:

1. **a)** The standard deviation of the weight of jam in a jar.
2. **b)** The proportion of jars where the contents weigh more than 115 grams.
3. **c)** A random sample of 25 jars is taken. Find the probability that 2 or fewer jars have contents weighing more than 115 grams.

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### Solution Approach:

Let's work through these parts step by step:

### Part (a): Finding the Standard Deviation

Given:
- Mean (\(\mu\)) = 108 grams.
- 3% of jars have contents weighing less than 100 grams.
- The distribution is normal.

To find the standard deviation (\(\sigma\)), we use the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
For 3% on the lower end of a normal distribution:
- \( X \) = 100 grams.
- 3% corresponds to a Z-score of approximately \(-1.88\).

Substitute into the formula:
\[
-1.88 = \frac{100 - 108}{\sigma}
\]
\[
\sigma = \frac{100 - 108}{-1.88} = \frac{-8}{-1.88} \approx 4.26 \text{ grams}
\]

### Part (b): Proportion of jars with weight > 115 grams

Now, we want to find the proportion of jars where \( X > 115 \).

- Mean (\(\mu\)) = 108 grams.
- Standard deviation (\(\sigma\)) = 4.26 grams.

Find the Z-score for \( X = 115 \):
\[
Z = \frac{115 - 108}{4.26} \approx 1.64
\]

Using standard normal tables, the probability for \( Z > 1.64 \) is approximately 0.05 (5%).

So, about 5% of jars will have contents weighing more than 115 grams.

### Part (c): Probability that 2 or fewer jars out of 25 weigh more than 115 grams

This is a binomial probability problem with:
- \( n = 25 \)
- \( p = 0.05 \) (probability that a jar weighs more than 115 grams)
- We need the probability that \( X \leq 2 \).

Using the binomial probability formula:
\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]
where
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

Let's calculate the probabilities:
1. \( P(X = 0) = \binom{25}{0} (0.05)^0 (0.95)^{25} \)
2. \( P(X = 1) = \binom{25}{1} (0.05)^1 (0.95)^{24} \)
3. \( P(X = 2) = \binom{25}{2} (0.05)^2 (0.95)^{23} \)

We'll need a calculator or software for precise numerical results for these terms, but the formula outlines the approach.

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Would you like further explanation on any part or help with the calculations? Here are some related questions for practice:

1. How do you interpret the Z-score in a normal distribution?
2. What is the significance of the normality assumption in this problem?
3. How does the value of the standard deviation affect the spread of the normal curve?
4. What is the difference between the binomial distribution and the normal approximation?
5. How would the answer change if the probability of jars weighing more than 115 grams was 10%?

**Tip:** When dealing with normal distributions, it's crucial to understand how to convert probabilities using Z-scores to standardize different values.

Jam is sold in jars, and the mean weight of the contents is 108 grams. Only 3% of jars have contents weighing less than 100 grams. Assuming that the weight of jam in a jar is normally distributed, find: a) the standard deviation of the weight of jam in a jar, b) the proportion of jars where the contents weigh more than 115 grams, c) A random sample of 25 jars is taken. Find the probability that 2 or fewer jars have contents weighing more than 115 grams.

Solve a problem involving normal and binomial distributions where the mean weight of jam jars is 108 grams, with a 3% probability of weighing less than 100 grams. Learn how to find the standard deviation, the proportion of jars weighing more than 115 grams, and the probability of 2 or fewer jars out of 25 exceeding 115 grams. Suitable for Grade 12 or early university students.

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