 A company producing orange juice buys all its oranges from a large orange orchard. The amount of juice that can be squeezed from each of these oranges is approximately normally distributed with a mean of 4.7 ounces and some unknown standard deviation. The company’s production manager knows that the probability is 30.85% that a randomly selected orange will contain less than 4.5 ounces of juice. Also, the probability is 10.56% that a randomly selected orange will contain more than 5.2 ounces of juice.
a) What is the probability that a randomly selected orange will contain between 4.2 and 4.9 ounces of juices?
b) What is the probability that a randomly selected orange will contain more than 4.2 ounces of juices?"  

We are given a normally distributed random variable \( X \) representing the amount of juice from an orange, with a mean \( \mu = 4.7 \) ounces and an unknown standard deviation \( \sigma \). To solve this problem, we will first need to find the standard deviation \( \sigma \) by using the probabilities and corresponding values provided.

### Step 1: Standardization of the normal distribution
The random variable \( X \) follows a normal distribution \( N(\mu = 4.7, \sigma) \), so we can transform \( X \) to the standard normal variable \( Z \) using the formula:

\[
Z = \frac{X - \mu}{\sigma}
\]

We know two probabilities:
- \( P(X < 4.5) = 0.3085 \)
- \( P(X > 5.2) = 0.1056 \)

These probabilities correspond to the cumulative probabilities for standard normal values of \( Z \), so we will use the standard normal table (or inverse Z-scores) to find the corresponding Z-values for 4.5 and 5.2 ounces.

### Step 2: Finding the Z-scores for the known probabilities
- From the cumulative standard normal distribution table, the Z-score that corresponds to \( P(Z < z) = 0.3085 \) is approximately \( z = -0.5 \). This gives us:
\[
Z = \frac{4.5 - 4.7}{\sigma} = -0.5
\]
Solving for \( \sigma \):
\[
\sigma = \frac{4.5 - 4.7}{-0.5} = \frac{-0.2}{-0.5} = 0.4
\]

- Now, to verify with the second probability \( P(X > 5.2) = 0.1056 \), we first note that \( P(X > 5.2) = 1 - P(X < 5.2) \), so:
\[
P(X < 5.2) = 1 - 0.1056 = 0.8944
\]
The Z-score corresponding to \( P(Z < z) = 0.8944 \) is approximately \( z = 1.25 \). Now check this with the calculated \( \sigma \):
\[
Z = \frac{5.2 - 4.7}{\sigma} = 1.25
\]
Solving for \( \sigma \):
\[
\sigma = \frac{5.2 - 4.7}{1.25} = \frac{0.5}{1.25} = 0.4
\]

Thus, the standard deviation is \( \sigma = 0.4 \).

### Step 3: Answering the questions

#### a) Probability that a randomly selected orange will contain between 4.2 and 4.9 ounces of juice
We want to find \( P(4.2 < X < 4.9) \). First, we convert these values to their corresponding Z-scores:
\[
Z_{4.2} = \frac{4.2 - 4.7}{0.4} = \frac{-0.5}{0.4} = -1.25
\]
\[
Z_{4.9} = \frac{4.9 - 4.7}{0.4} = \frac{0.2}{0.4} = 0.5
\]

Now we use the standard normal table to find the probabilities:
- \( P(Z < -1.25) \approx 0.1056 \)
- \( P(Z < 0.5) \approx 0.6915 \)

Thus, the probability that the juice amount is between 4.2 and 4.9 ounces is:
\[
P(4.2 < X < 4.9) = P(Z < 0.5) - P(Z < -1.25) = 0.6915 - 0.1056 = 0.5859
\]

#### b) Probability that a randomly selected orange will contain more than 4.2 ounces of juice
We want to find \( P(X > 4.2) \). This is the complement of \( P(X < 4.2) \):
\[
P(X > 4.2) = 1 - P(X < 4.2)
\]

From earlier, we know \( P(X < 4.2) = P(Z < -1.25) \approx 0.1056 \), so:
\[
P(X > 4.2) = 1 - 0.1056 = 0.8944
\]

### Final Answers:
- a) The probability that a randomly selected orange will contain between 4.2 and 4.9 ounces of juice is approximately **0.5859**.
- b) The probability that a randomly selected orange will contain more than 4.2 ounces of juice is approximately **0.8944**.

Would you like more details or have any questions? Here are 5 related questions to consider:

1. How can we use the normal distribution to approximate other types of distributions?
2. What happens to the probabilities if the standard deviation of juice yield increases?
3. How would the calculation change if the mean juice yield was different?
4. Can you find the probability for a different range, such as between 4.5 and 5.5 ounces?
5. How does the Central Limit Theorem apply to the production of orange juice in this case?

**Tip:** Remember that cumulative normal distribution tables give the area to the left of a Z-score. Use complements for probabilities to the right.

A company producing orange juice buys all its oranges from a large orange orchard. The amount of juice that can be squeezed from each of these oranges is approximately normally distributed with a mean of 4.7 ounces and some unknown standard deviation. The company’s production manager knows that the probability is 30.85% that a randomly selected orange will contain less than 4.5 ounces of juice. Also, the probability is 10.56% that a randomly selected orange will contain more than 5.2 ounces of juice. a) What is the probability that a randomly selected orange will contain between 4.2 and 4.9 ounces of juices? b) What is the probability that a randomly selected orange will contain more than 4.2 ounces of juices?

Learn how to calculate probabilities for juice yields from oranges using normal distribution. With a mean of 4.7 ounces, the problem explores finding the probability of juice amounts between 4.2 and 4.9 ounces and more than 4.2 ounces. Step-by-step solution provided with explanations of Z-scores and standard deviations. Suitable for Grades 11-12.

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