Mathematics: Mr. smith s first test s data: 55, 42, 78, 99, 69, 83, 74, 83, 97. what is the value of q1? (remember, the median for the four numbers, 42, 55, 69, and 74, is between 55 and 69. to find the median between two numbers, you take the average of those two numbers.)

Mr. smith s first test s data: 55, 42, 78, 99, 69,

Ответ:

efloyd04

17.02.2020

Since n(1-p) = 3.6 < 5 , the normal distribution cannot be used as an approximation to the binomial probability to approximate the probability.

Using the binomial distribution, 100% probability that more than 97 out of 120 people will get the flu this winter.

Step-by-step explanation:

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.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume the probability that a given person will get the flu this winter is 97%.

This means that p = 0.97

120 people

This means that n = 120

Verifying the necessary conditions.

np = 120*0.97 = 116.4

n(1-p) = 120*0.03 = 3.6

Since n(1-p) = 3.6 < 5 , the normal distribution cannot be used as an approximation to the binomial probability to approximate the probability. Thus, the binomial distribution has to be used.

Probability using the binomial distribution:

P(X 97) = P(X = 98) + ... + P(X = 120)

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

$P(X = 98) = C_{120,98}.(0.97)^{98}.(0.03)^{22} \approx 0$

Probability close to 0, but below the mean, which means that the probability of the number being above this is 100%.

Using the binomial distribution, 100% probability that more than 97 out of 120 people will get the flu this winter.

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Mr. smith's first test's data: 55, 42, 78, 99, 69, 83, 74, 83, 97. what is the value of q1? (remember, the median for the four numbers, 42, 55, 69, and 74, is between 55 and 69. to find the median between two numbers, you take the average of those two numbers.)

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