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Ivenika [448]
2 years ago
7

Help ASAP! Explain the process please

Mathematics
1 answer:
Len [333]2 years ago
4 0

Answer:

B. 650 Roast Beef, 490 Tuna, 900 Turkey

Step-by-step explanation:

As this chart displays the order that occured in one week, in order to find the estimate for the month is to multiply the values on the chart by how many weeks there are in a month.

As there are 4 weeks in a months, we can multiply each value by this ratio

Roast Beef: 4*152=608

Tuna: 4*114=456

Turkey: 4*209=836

As this is an estimate, we are looking for the option that is the closest to these estimated values.

Option B is the closest to these values.

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Step-by-step explanation:

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The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2617 and standard deviation
Eduardwww [97]

Answer:

a) N(2617, 586)

b) 0.3613 = 36.13% probability that the customer consumes less than 2409 calories.

c) 0.4013 = 40.13% of the customers consume over 2764 calories

d) 3981 calories.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 2617, \sigma = 586

a. What is the distribution of X?

Here we first place the mean, then the standard deviation.

N(2617, 586)

b. Find the probability that the customer consumes less than 2409 calories.

This is the pvalue of Z when X = 2409. So

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

Z = \frac{2409 - 2617}{586}

Z = -0.355

Z = -0.355 has a pvalue of 0.3613

0.3613 = 36.13% probability that the customer consumes less than 2409 calories.

c. What proportion of the customers consume over 2764 calories?

This is 1 subtracted by the pvalue of Z when X = 2764. So

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

Z = \frac{2764 - 2617}{586}

Z = 0.25

Z = 0.25 has a pvalue of 0.5987

1 - 0.5987 = 0.4013

0.4013 = 40.13% of the customers consume over 2764 calories

d. The Piggy award will given out to the 1% of customers who consume the most calories. What is the fewest number of calories a person must consume to receive the Piggy award?

Top 1%, so the 100-1 = 99th percentile.

The 99th percentile is the value of X when Z has a pvalue of 0.99. So it is X when Z = 2.327. So

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

2.327 = \frac{X - 2617}{586}

X - 2617 = 2.327*586

X = 3980.6

Rounding to the nearest calorie, 3981 calories.

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