Question

A population of values has a normal distribution with u = 203.6 and o = 35.5. You intend to draw a randomsample of size n = 16.Find the probability that a single randomly selected value is greater than 231.1.PIX > 231.1) =Find the probability that a sample of size n = 16 is randomly selected with a mean greater than 231.1.P(M > 231.1) =Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or Z-scores rounded to 3 decimal places are accepted.

Answer 1

Part 1:

The probability that a single randomly selected value is greater than 231.1 equals one minus the probability that it is less or equal to 231.1:

P(x > 231.1) = 1 - P(x ≤ 231.1)

Now, to find P(x ≤ 231.1), we can transform x in its correspondent z-score, and then use a z-score table to find the probability:

x ≤ 231.1 => z ≤ (231.1 - 203.6)/35.5, because z = (x - mean)/(standard deviation)

z ≤ 0.775 (rounding to 3 decimal places)

Then we have:

P(x ≤ 231.1) = P( z ≤ 0.775)

Now, using a table, we find:

P( z ≤ 0.775) ≅ 0.7808

Then, we have:

P(x > 231.1) ≅ 1 - 0.7808 = 0.2192

Therefore, the asked probability is approximately 0.2192.

Part 2

For the next part, since we will select a sample out of other samples with size n = 16, we need to use the formula:

z = (x - mean)/(standard deviation/√n)

Now, x represents the mean of the selected sample, which we want to be greater than 231.1. Then, we have:

z = (231.1 - 203.6)/(35.5/√16) = 27.5/(35.5/4) = 3.099

P(x > 231.1) = 1 - P(x ≤ 231.1) = 1 - P(x ≤ 231.1) = 1 - P( z ≤ 3.099) = 1 - 0.9990 = 0.0010

Therefore, the asked probability is approximately 0.0010.