Question

Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about 6.9. Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean to answer the following questions. (a) Would it be unusual to have more than 120 successes out of 200 trials

Answer 1

Answer:

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

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 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.

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

In this problem, we have that:

$\mu = 80, \sigma = 6.9$

Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean

This means that z-scores higher than 2.5 or lower than -2.5 are considered unusual.

Would it be unusual to have more than 120 successes out of 200 trials

We have to find the Z-score of X = 120.

So

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

$Z = \frac{120 - 80}{6.9}$

$Z = 5.8$

120 has a z-score higher than 2.5. So yes, it would be unusual to have more than 120 successes out of 200 trials.