Let’s work through each part of the question step-by-step.

Part (a)

We want to calculate the percentage of apples that Jane does not like, which are those with weights less than or equal to 100 grams. Since the weights of the apples are normally distributed with a mean ($\mu$) of 180g and a standard deviation ($\sigma$) of 60g, we’ll use the Z-score formula to standardize 100g:

$Z = \frac{X - \mu}{\sigma}$

Calculating the Z-score for 100g:

$Z = \frac{100 - 180}{60} = \frac{-80}{60} = -1.33$

Using a standard normal distribution table, we find the probability that a Z-score is less than or equal to -1.33. The Z-table gives us:

$P(Z \leq -1.33) \approx 0.0918$

So, about 9.18% of the apples that Jane picks weigh 100g or less, which are the ones she doesn’t like.

Part (b)

Let $q_3$ represent the minimum weight for the upper quartile (top 25%) of apples liked by Jane. To find $q_3$, we need the Z-score that corresponds to the 75th percentile (since only 25% of weights will be greater than or equal to $q_3$).

From the Z-table, the Z-score corresponding to the 75th percentile is approximately $Z = 0.675$.

Finding $q_3$ (the weight that corresponds to the 75th percentile):

$q_3 = \mu + Z \times \sigma$ $q_3 = 180 + 0.675 \times 60 = 180 + 40.5 = 220.5 \text{ grams}$

Probability that the weight of a randomly selected apple is $\geq q_3$:

Since we are looking at only the apples that Jane likes (weight $> 100 \text{g}$), and within that group, we want the probability that the apple weighs at least $q_3 = 220.5$ grams.

1. Step 1: Probability that an apple weighs more than 100g (apples Jane likes): $P(X > 100) = 1 - P(X \leq 100) = 1 - 0.0918 = 0.9082$

2. Step 2: Probability that an apple weighs at least $q_3$ (220.5g) given that Jane likes it: $P(X \geq 220.5 \mid X > 100) = \frac{P(X \geq 220.5)}{P(X > 100)}$

   We need $P(X \geq 220.5)$. For $X = 220.5$: $Z = \frac{220.5 - 180}{60} = \frac{40.5}{60} = 0.675$

   From the Z-table: $P(Z \leq 0.675) \approx 0.75 \Rightarrow P(Z \geq 0.675) = 1 - 0.75 = 0.25$

3. Final Calculation: $P(X \geq 220.5 \mid X > 100) = \frac{0.25}{0.9082} \approx 0.2753$

Therefore, the probability that a randomly selected apple that Jane likes (weighing more than 100g) weighs at least $q_3 = 220.5$ grams is approximately 27.53%.

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Summary

1. (a) The percentage of apples that Jane does not like is 9.18%.
2. (b) The probability that an apple Jane likes weighs at least $q_3 = 220.5$ grams is approximately 27.53%.

Would you like more details on any of these steps?

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Here are 5 related questions for further exploration:

1. What is the probability that an apple weighs between 100g and 180g?
2. How would the results change if the standard deviation was increased to 80g?
3. What is the probability that an apple weighs between 180g and 220.5g?
4. How does changing the mean to 200g affect the percentage of apples Jane dislikes?
5. What weight corresponds to the 90th percentile of the apple distribution?

Tip: When using Z-scores, always check that the distribution is normal; otherwise, the Z-table method may not apply directly.