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Sagot :
Answer:
the probability = 0.3594
Step-by-step explanation:
We can find the probability that a randomly selected slice has more than 200 calories by first finding its Z-score:
[tex]\boxed{Z=\frac{x-\mu}{\sigma} }[/tex]
where:
- [tex]Z=\text{Z-score}[/tex]
- [tex]x=\text{observed value}[/tex]
- [tex]\mu=\text{mean}[/tex]
- [tex]\sigma=\text{standard deviation}[/tex]
Given:
- [tex]\mu=191[/tex]
- [tex]\sigma=25[/tex]
- [tex]x=200[/tex]
Then:
[tex]\begin{aligned}Z&=\frac{x-\mu}{\sigma} \\\\&=\frac{200-191}{25}\\\\&=0.36 \end{aligned}[/tex]
With the normal distribution table (refer to the attached picture), we can find the probability that:
[tex]P(Z \leq 0.36)=0.6406[/tex]
Since we are looking for the probability of P(Z > 0.36), then:
[tex]\begin{aligned}P(Z > 0.36)&=1-P(Z\leq 0.36)\\&=1-0.6406\\&=\bf 0.3594 \end{aligned}[/tex]
The probability that a slice of large pepperoni pizza contains more than 200 calories is approximately 35.94%.
The problem states that the amount of calories in a slice of large pepperoni pizza is normally distributed with a mean (μ) of 191 and a standard deviation (σ) of 25. We need to find the probability that a randomly selected slice has more than 200 calories.
First, we convert this normal distribution problem into a standard normal distribution (Z) problem using the Z-score formula:
Z = (X - μ) / σ
Here, X is 200, the value for which we are finding the probability:
Z = (200 - 191) / 25
Z = 9 / 25
Z = 0.36
Next, we use the Z-score to find the probability. The Z-score of 0.36 corresponds to a probability of 0.6406 (from Z-tables or using a normal distribution calculator), which represents the probability that a slice has fewer than 200 calories.
To find the probability that a slice has more than 200 calories, we subtract this value from 1:
P(X > 200) = 1 - P(X < 200)
P(X > 200) = 1 - 0.6406
P(X > 200) = 0.3594
Therefore, the probability that a randomly selected slice of large pepperoni pizza has more than 200 calories is approximately 0.3594, or 35.94%.
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