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mina [271]
3 years ago
12

find the amount each contestant would carry if the total amount carried (5 cups) was redistributed equally among all the contest

ants.

Mathematics
1 answer:
Blizzard [7]3 years ago
3 0

<u>Complete Question</u>

The line plot shows the amount of green ooze (in cup) carried by ten contestants. Find the amount each contestant would carry if the total amount carried (5 cups) was redistributed equally among all the contestants.

Answer:

1/2 Cup

Step-by-step explanation:

This is a Line Plot in which:

  • 1/4 had two Xs
  • 3/8 had two Xs
  • 1/2 had two Xs
  • 5/8 had three Xs
  • 7/8 had one Xs

Total

=(\frac{1}{4}X2) +(\frac{3}{8}X2) +(\frac{1}{2}X2) +(\frac{5}{8}X3) +(\frac{7}{8}X1) \\\\=\frac{2}{4} +\frac{6}{8} +1 +\frac{15}{8} +\frac{7}{8}\\\\$=5 Cups

If this amount was redistributed equally among the ten contestants:

Each contestant would carry (5/10) cups =1/2 cup.

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Answer:

D. 0.9938.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 115 and a standard deviation of 8.

This means that \mu = 115, \sigma = 8

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This means that n = 100, s = \frac{8}{\sqrt{100}} = 0.8

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This is the p-value of Z when X = 117, so:

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

By the Central Limit Theorem

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

Z = \frac{117 - 115}{0.8}

Z = 2.5

Z = 2.5 has a p-value of 0.9938, and thus, the correct answer is given by option D.

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