Question

a population of values has a normal distribution with $\mu = 203.1$ and $\sigma = 50.5$.

a. find the probability that a single randomly selected value is less than 208.6. round your answer to four decimal places.
$p(x < 208.6) =$

b. find the probability that a randomly selected sample of size $n = 187$ has a mean less than 208.6. round your answer to four decimal places.
$p(m < 208.6) =$

Explanation:

Step1: Calculate the z - score for part a

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(x = 208.6\), \(\mu=203.1\), and \(\sigma = 50.5\).

$$z=\frac{208.6 - 203.1}{50.5}=\frac{5.5}{50.5}\approx0.109$$

Step2: Find the probability for part a

Using the standard normal distribution table or a calculator with a normal - distribution function (\(P(Z<z)\)), for \(z = 0.109\), \(P(X<208.6)=P(Z < 0.109)\approx0.5438\)

Step3: Calculate the standard error for part b

The formula for the standard error is \(\sigma_M=\frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = 50.5\) and \(n = 187\)
\(\sigma_M=\frac{50.5}{\sqrt{187}}\approx\frac{50.5}{13.67}\approx3.694\)

Step4: Calculate the z - score for part b

The formula for the z - score for the sample mean is \(z=\frac{M-\mu}{\sigma_M}\). Here, \(M = 208.6\), \(\mu = 203.1\), and \(\sigma_M\approx3.694\)

$$z=\frac{208.6-203.1}{3.694}=\frac{5.5}{3.694}\approx1.49$$

Step5: Find the probability for part b

Using the standard normal distribution table or a calculator with a normal - distribution function (\(P(Z<z)\)), for \(z = 1.49\), \(P(M<208.6)=P(Z < 1.49)\approx0.9319\)

Answer:

a. \(0.5438\)
b. \(0.9319\)