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Semmy [17]
4 years ago
7

Wyatt has a bag that contains pineapple chews, cherry chews, and lime chews. He performs an experiment. Wyatt randomly removes a

chew from the bag, records the result, and returns the chew to the bag. Wyatt performs the experiment 33 times. The results are shown below: A pineapple chew was selected 23 times. A cherry chew was selected 6 times. A lime chew was selected 4 times. Based on these results, express the probability that the next chew Wyatt removes from the bag will be lime chew as a percent to the nearest whole number.
Mathematics
1 answer:
arlik [135]4 years ago
5 0

The possibility Wyatt has of getting a lime chew would be 12% because his chance would be according to his experiments would be 4/33, which means that the percentage of Wyatt getting the lime chew would be 12.1212% (which the neares whole number percentage would be 12%).

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Answer:

8.81% probability that the student answers exactly 4 questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A multiple choice exam has ten questions.

This means that n = 10

The probability of answering any question correctly is 0.20.

This means that p = 0.2

What is the probability that the student answers exactly 4 questions correctly

This is P(X = 4).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{10,4}.(0.2)^{4}.(0.8)^{6} = 0.0881

8.81% probability that the student answers exactly 4 questions correctly

4 0
4 years ago
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds,What perc
soldier1979 [14.2K]

Answer:

57.62% of players weigh between 180 and 220 pounds

Step-by-step explanation:

When the distribution is normal, we use 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 = 200, \sigma = 25

What percent of players weigh between 180 and 220 pounds

We have to find the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 180.

X = 220

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

Z = \frac{220 - 200}{25}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881

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

Z = \frac{180 - 200}{25}

Z = -0.8

Z = -0.8 has a pvalue of 0.2119

0.7881 - 0.2119 = 0.5762

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Answer:

see explanation

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