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djverab [1.8K]
3 years ago
10

PLEASE HELP ME i'll give 100 points and brainliest to who comes up with the best answer!!!! Collect data on a chance event of yo

ur choice with a calculable theoretical probability. Use the relative frequency to estimate the theoretical probability of that event. Compare the experimental and theoretical probability and discuss the differences between them.
Mathematics
2 answers:
dezoksy [38]3 years ago
8 0

Ima just say 124,098,868 am i close?

Damm [24]3 years ago
8 0

Correct. The theoretical probability of rolling a 2 is 1/4 = 25%, while Joan's experience has the relative frequency at 187/2,000 = 9.35%, which is quite different. Joan's experience includes a large number of trials so the Law of Large Numbers would suggest that the theoretical and empirical probabilities should be similar. Therefore, her experience does call into question the die as possibly not being fair.

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Assume that the population of human body temperatures has a mean of 98.6 degrees F and a standard deviation of 0.62 degrees F. I
dimulka [17.4K]

Answer:

0% probability of getting a mean temperature of 98.2 degrees F or lower.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

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 problem, we have that:

\mu = 98.6, \sigma = 0.62, n = 106, s = \frac{0.62}{\sqrt{106}} = 0.06

Find the probability of getting a mean temperature of 98.2 degrees F or lower.

This is the pvalue of Z when X = 98.2. So

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

Z = \frac{98.2 - 98.6}{0.06}

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Z = -6.67 has a pvalue of 0.

So there is a 0% probability of getting a mean temperature of 98.2 degrees F or lower.

8 0
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