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Katarina [22]
3 years ago
7

A number cube was rolled as part of an experiment. The results are shown in the table. Explain how to find the experimental prob

ability of rolling a 6. A 2-column table with 6 rows. Column 1 is labeled Number with entries 1, 2, 3, 4, 5, 6. Column 2 is labeled Frequency with entries 13, 11, 9, 8, 10, 9.
Mathematics
2 answers:
pychu [463]3 years ago
4 0

The experimental probability of rolling a 6 is 9/60  which can be determined by dividing the frequency of the observation 6 with the total frequency of the experiment.

<u>Step-by-step explanation:</u>

Experimental probability is different from theoretical probability because the former is obtained by experimentation  while the latter is what we expect theoretically.When we take a number of observations, the experimental probability and theoretical probability need not be the same.

In this question we have to determine the experimental probability of 6. It can be determined by dividing the frequency of the observation 6 by the total frequency of the experiment.

frequency of 6=9

total frequency=frequency of 1+frequency of 2+frequency of 3+frequency of 4+frequency of 5+frequency of 6

=13+11+9+8+10+9

=60

P(6)=frequency of 6/total frequency

=9/60

jolli1 [7]3 years ago
3 0

Answer:

9/60

Step-by-step explanation:

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The annual rainfall (in inches) in a certain region is normally distributed with mean 43.2 and variance 20.8. Assume rainfall in
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Answer:

92.24% probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the binomial probability distribution.

Normal probability distribution:

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

In a set with mean \mu and standard deviation(which is the square root of the variance) \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.

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.

In this problem, we have that:

\mu = 43.2, \sigma = \sqrt{20.8} = 4.56

Probability that a year has rainfall of more than 50 inches.

pvalue of Z when X = 50. So

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

Z = \frac{50 - 43.2}{4.56}

Z = 1.49

Z = 1.49 has a pvalue of 0.9319

1 - 0.9319 = 0.0681

Find the probability that out of 15 years, at most 2 have rainfall of more than 50 inches.

This is P(X \leq 2) when n = 15, p = 0.0681. So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

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

P(X = 0) = C_{15,0}.(0.0681)^{0}.(0.9319)^{15} = 0.3472

P(X = 1) = C_{15,1}.(0.0681)^{1}.(0.9319)^{14} = 0.3805

P(X = 2) = C_{15,2}.(0.0681)^{2}.(0.9319)^{13} = 0.1947

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3472 + 0.3805 + 0.1947 = 0.9224

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