Question
Tofurkey is a vegan turkey substitute, usually made from tofu. At a certain restaurant, the number of calories in a serving of tofurkey with wild mushroom stuffing and gravy is normally distributed with mean 477 and standard deviation 30.
(a) What proportion of servings have between 481 and 535 calories? Round the answer to four decimal places.
(a) What proportion of servings have between 481 and 535 calories? Round the answer to four decimal places.
Answers
Answer
(b) Find the 75th percentile of the number of calories. Round the answer to two decimal places.
Answered by
GPT-4o mini
To find the proportion of servings of tofurkey that have between 481 and 535 calories, we will use the properties of the normal distribution. We know that the calories are normally distributed with a mean (μ) of 477 and a standard deviation (σ) of 30.
First, we need to convert the calorie values (481 and 535) into z-scores using the formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where \(X\) is the value we want to convert to a z-score.
1. **Calculate the z-score for 481 calories**:
\[
z_{481} = \frac{(481 - 477)}{30} = \frac{4}{30} \approx 0.1333
\]
2. **Calculate the z-score for 535 calories**:
\[
z_{535} = \frac{(535 - 477)}{30} = \frac{58}{30} \approx 1.9333
\]
Next, we will look up these z-scores in the standard normal distribution table or use a calculator to find the corresponding probabilities.
- For \(z_{481} \approx 0.1333\):
Using standard normal distribution tables or a calculator, we find:
\[
P(Z < 0.1333) \approx 0.5534
\]
- For \(z_{535} \approx 1.9333\):
Again, using standard normal distribution tables or a calculator, we find:
\[
P(Z < 1.9333) \approx 0.9734
\]
Now, to find the proportion of servings between 481 and 535 calories, we subtract the cumulative probability at \(z_{481}\) from the cumulative probability at \(z_{535}\):
\[
P(481 < X < 535) = P(Z < 1.9333) - P(Z < 0.1333)
\]
\[
P(481 < X < 535) = 0.9734 - 0.5534 = 0.42
\]
Therefore, the proportion of servings that have between 481 and 535 calories is:
\[
\boxed{0.4200}
\]
First, we need to convert the calorie values (481 and 535) into z-scores using the formula:
\[
z = \frac{(X - \mu)}{\sigma}
\]
where \(X\) is the value we want to convert to a z-score.
1. **Calculate the z-score for 481 calories**:
\[
z_{481} = \frac{(481 - 477)}{30} = \frac{4}{30} \approx 0.1333
\]
2. **Calculate the z-score for 535 calories**:
\[
z_{535} = \frac{(535 - 477)}{30} = \frac{58}{30} \approx 1.9333
\]
Next, we will look up these z-scores in the standard normal distribution table or use a calculator to find the corresponding probabilities.
- For \(z_{481} \approx 0.1333\):
Using standard normal distribution tables or a calculator, we find:
\[
P(Z < 0.1333) \approx 0.5534
\]
- For \(z_{535} \approx 1.9333\):
Again, using standard normal distribution tables or a calculator, we find:
\[
P(Z < 1.9333) \approx 0.9734
\]
Now, to find the proportion of servings between 481 and 535 calories, we subtract the cumulative probability at \(z_{481}\) from the cumulative probability at \(z_{535}\):
\[
P(481 < X < 535) = P(Z < 1.9333) - P(Z < 0.1333)
\]
\[
P(481 < X < 535) = 0.9734 - 0.5534 = 0.42
\]
Therefore, the proportion of servings that have between 481 and 535 calories is:
\[
\boxed{0.4200}
\]
Answered by
GPT-4o mini
To find the 75th percentile of the number of calories in servings of tofurkey, we need to determine the z-score that corresponds to the 75th percentile in the standard normal distribution.
Using standard normal distribution tables or a calculator, the z-score that corresponds to the 75th percentile is approximately:
\[
z \approx 0.6745
\]
Now, we can use the z-score to find the corresponding calorie value (X) using the following formula:
\[
X = \mu + z \cdot \sigma
\]
Where:
- \(\mu\) is the mean (477 calories),
- \(z\) is the z-score (0.6745),
- \(\sigma\) is the standard deviation (30 calories).
Plugging in the values:
\[
X = 477 + 0.6745 \cdot 30
\]
\[
X = 477 + 20.235 \approx 497.235
\]
Rounding to two decimal places, the 75th percentile of the number of calories is:
\[
\boxed{497.24}
\]
Using standard normal distribution tables or a calculator, the z-score that corresponds to the 75th percentile is approximately:
\[
z \approx 0.6745
\]
Now, we can use the z-score to find the corresponding calorie value (X) using the following formula:
\[
X = \mu + z \cdot \sigma
\]
Where:
- \(\mu\) is the mean (477 calories),
- \(z\) is the z-score (0.6745),
- \(\sigma\) is the standard deviation (30 calories).
Plugging in the values:
\[
X = 477 + 0.6745 \cdot 30
\]
\[
X = 477 + 20.235 \approx 497.235
\]
Rounding to two decimal places, the 75th percentile of the number of calories is:
\[
\boxed{497.24}
\]
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