Question

a population of values has a normal distribution with \\( \mu = 8.2 \\) and \\( \sigma = 40.5 \\). you intend to draw a random sample of size \\( n = 22 \\).
find the probability that a single randomly selected value is less than -1.3.
\\( p(x < -1.3)= \\)
find the probability that a sample of size \\( n = 22 \\) is randomly selected with a mean less than -1.3.
\\( p(m < -1.3)= \\)
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: video post to forum

Explanation:

Step1: Calculate z - score for single value

The formula for z - score is \(z=\frac{x-\mu}{\sigma}\).
Given \(x = - 1.3\), \(\mu=8.2\), \(\sigma = 40.5\)
\(z=\frac{-1.3 - 8.2}{40.5}=\frac{-9.5}{40.5}\approx - 0.235\)

Step2: Find probability for single value

Using the standard normal distribution table (or a calculator with a normal - distribution function), \(P(X\lt - 1.3)=P(Z\lt - 0.235)\)
\(P(Z\lt - 0.235)=0.4075\) (using a calculator: \(\text{normalcdf}(-\infty,-0.235)\))

Step3: Calculate z - score for sample mean

The formula for the z - score of the sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\)
Given \(\bar{x}=-1.3\), \(\mu = 8.2\), \(\sigma = 40.5\), \(n = 22\)
\(\frac{\sigma}{\sqrt{n}}=\frac{40.5}{\sqrt{22}}\approx\frac{40.5}{4.690}\approx8.635\)
\(z=\frac{-1.3 - 8.2}{8.635}=\frac{-9.5}{8.635}\approx - 1.100\)

Step4: Find probability for sample mean

Using the standard normal distribution table (or a calculator with a normal - distribution function), \(P(M\lt - 1.3)=P(Z\lt - 1.100)\)
\(P(Z\lt - 1.100)=0.1357\) (using a calculator: \(\text{normalcdf}(-\infty,-1.100)\))

Answer:

\(P(X\lt - 1.3)=0.4075\)
\(P(M\lt - 1.3)=0.1357\)