Question

the amount of calories consumed by customers at the chinese buffet is normally distributed with mean 2885 and standard deviation 651.
a. find the probability that the customer consumes less than 2000 calories.
select

b. what percent of the customers consume over 4,200 calories?
select

c. how many will be expected to eat more than 3000 if 287 people are dining at the buffet? select

Explanation:

Step1: Calculate z-score for 2000

$z = \frac{X - \mu}{\sigma} = \frac{2000 - 2885}{651} \approx -1.36$

Step2: Find P(Z < -1.36)

Using standard normal table: $P(Z < -1.36) \approx 0.0869$
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Step3: Calculate z-score for 4200

$z = \frac{4200 - 2885}{651} \approx 2.02$

Step4: Find P(Z > 2.02)

$P(Z > 2.02) = 1 - P(Z < 2.02) \approx 1 - 0.9783 = 0.0217$, or 2.17%
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Step5: Calculate z-score for 3000

$z = \frac{3000 - 2885}{651} \approx 0.18$

Step6: Find P(Z > 0.18)

$P(Z > 0.18) = 1 - P(Z < 0.18) \approx 1 - 0.5714 = 0.4286$

Step7: Calculate expected number

$\text{Expected number} = 287 \times 0.4286 \approx 123$

Answer:

A. 0.0869
B. 2.17%
C. 123