Question

A population of values has a normal distribution with u = 223 and o = 21. You intend to draw a randomsample of size n = 22.Find the probability that a single randomly selected value is less than 234.2.P(X < 234.2) =Find the probability that a sample of size n = 22 is randomly selected with a mean less than 234.2.P(M < 234.2) =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.Question Help: D VideoSubmit Question

Answer 1

Population mean μ = 223

Population standard deviation σ = 21

Sample size n = 22

What to find:

x < 234.2

mean less than 234.2

To get the probability of a random value from a set of values, we use the formula below:

`z=(x-\mu)/(\sigma)`

From the formula, we will substitute x with 234.2, μ with 223, and σ with 21 based on the given information.

`z=(234.2-223)/(21)=(11.2)/(21)\approx0.53`

The z equivalent of 234.2 is 0.53. Looking at the z- table, the area less than 0.53 covers 0.7019 of the normal curve. Therefore, the probability that a single randomly selected value is less than 234.2 is 0.7019 or 70.19%.

Though the first question and second questions look the same, well, they are not. The first question is looking for a random value x. The second question is looking for a random mean. If we are looking for a random mean, we'll have to modify the formula to be used. This time, we will be using this formula:

`z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}}`

where bar x will be the mean 234.2, μ = 223, σ = 21, and n = 22 based on the given information.

`z=\frac{234.2-223}{\frac{21}{\sqrt[]{22}}}=(11.2)/(4.477215043)\approx2.50`

Looking at the z-table, the area less than z = 2.50 is 0.9938 therefore, the probability that a sample of size n = 22 is randomly selected with a mean less than 234.2 is 0.9938 or 99.38%.