Mathematics, 29.05.2020 14:57 clmorcutt420

Suppose 30% of a population possess a given characteristic. If a random sample of size 1200 is drawn from the population, then the probability that less than 348 possess that characteristic is

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Answer from: orlando19882000

The population follows a normal distribution.

The probability that less than 348 possess that characteristic is 0.2248

The given parameters are:

$\mathbf{n = 1200}$

$\mathbf{p = 30\%}$

Start by calculating the mean:

$\mathbf{\mu =np}$

$\mathbf{\mu =1200 \times 30\%}$

$\mathbf{\mu =360}$

Calculate the standard deviation

$\mathbf{\sigma = \sqrt{np(1 - p)}}$

$\mathbf{\sigma = \sqrt{360(1 - 30\%)}}$

$\mathbf{\sigma = \sqrt{252}}$

$\mathbf{\sigma = 15.87}$

Calculate the z-score

$\mathbf{z = \frac{x - \mu}{\sigma}}$

Where:

x = 348

So, we have:

$\mathbf{z = \frac{348 - 360}{15.87}}$

$\mathbf{z = -\frac{12}{15.87}}$

$\mathbf{z = -0.7561}$

So, the probability is represented as:

$\mathbf{P(x < 348) = P(z < -0.7561)}$

From the z-table of probabilities, we have:

$\mathbf{P(x < 348) = 0.2248}$

Hence, the probability that less than 348 possess that characteristic is 0.2248

Read more about probabilities at:

link

Answer from: DanteMadrid33

The probability that less than 348 possess that characteristic is 0.2148 = 21.48%.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .