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po3 Smith is eligible to take the po2 advancement exam the only evaluation he received his current pay greater than Frost evalua

andreev551 [17]

In the case above, Smith does not need anything, or nothing is required to occur to establish a PMA.

<h3>What is an advancement exam?</h3>

An advancement exam is known to be one that gives an unbiased factor in regards to the Final Multiple Score (FMS) algorithm and it is said to help in the rank order of qualified candidates in terms of advancement consideration.

Hence, In the case above, Smith does not need anything, or nothing is required to occur to establish a PMA.

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po3 Smith is elligible to take the po2 advancement exam. The only evaluation he received in his current paygrade is a frocked evaluation. what, if anything, is required to occur to establish a pma.

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2 years ago

In a recent year, about 22% of Americans 18 years and older are single. What is the probability that in a random sample of 200 A

Pie

Using the normal approximation to the binomial, it is found that there is a 0.0107 = 1.07% probability that more than 30 are single.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:

$\mu = np = 200(0.22) = 44$

$\sigma = \sqrt{np(1 - p)} = \sqrt{200(0.22)(0.78)} = 5.8583$

The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is <u>1 subtracted by the p-value of Z when X = 30.5</u>, hence:

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

$Z = \frac{30.5 - 44}{5.8583}$