Question

Question 6 The mineral content of a particular brand of supplement pills is normally distributed with mean 490 mg and variance of 400. What is the probability that a randomly selected pill contains at least 500 mg of minerals

Answer 1

Answer:

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean 490 mg and variance of 400.

This means that $\mu = 490, \sigma = \sqrt{400} = 20$

What is the probability that a randomly selected pill contains at least 500 mg of minerals?

This is 1 subtracted by the p-value of Z when X = 500. So

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

$Z = \frac{500 - 490}{20}$

$Z = 0.5$

$Z = 0.5$ has a p-value of 0.6915.

1 - 0.6915 = 0.3085

0.3085 = 30.85% probability that a randomly selected pill contains at least 500 mg of minerals

Answer 2

The probability is 0.3085.

Let X=the amount of minerals.

We have to find P(X≥500).

Mean = 490.

Standard deviation = $\sqrt{400}=20$.

z score of 500 is = $\frac{500-490}{20 } =0.5$.

So, P(X≥500) = P(z≥0.5) = 1 - P(z<0.5) = 1-0.6915=0.3085.

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