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Ostrovityanka [42]
3 years ago
10

A coin is tossed 20 times with the following results:

Mathematics
1 answer:
Rudik [331]3 years ago
4 0

Using it's concept, it is found that there is a 0.45 = 45% experimental probability of the coin landing on tails.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

For an experimental probability, these number of outcomes are taken from previous trials.

In this problem, 9 out of 20 trials resulted in tails, hence the experimental probability is given by:

p = 9/20 = 0.45 = 45%.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

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D + 4/3 = -1/3<br> find d.
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The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a stan
IgorLugansk [536]

Answer:

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

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 normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this problem, we have that:

\mu = 3.02, \sigma = 0.29, n = 20, s = \frac{0.29}{\sqrt{20}} = 0.0648

Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

This is 1 subtracted by the palue of Z when X = 3.10. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 3.02}{0.0648}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907

1 - 0.8907 = 0.1093

10.93% probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

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