Question

a population of values has a normal distribution with $mu = 65.3$ and $sigma = 48.6$. you intend to draw a random sample of size $n = 125$. please show your answers as numbers accurate to 4 decimal places. find the probability that a single randomly selected value is less than 62.7. $p(x < 62.7)=$ find the probability that a sample of size $n = 125$ is randomly selected with a mean less than 62.7. $p(\bar{x} < 62.7)=$

Explanation:

Step1: Calculate z - score for single value

The formula for the z - score of a single value is $z=\frac{x-\mu}{\sigma}$. Here, $x = 62.7$, $\mu=65.3$, and $\sigma = 48.6$. So, $z=\frac{62.7 - 65.3}{48.6}=\frac{- 2.6}{48.6}\approx - 0.0535$.

Step2: Find the probability for single value

We want $P(X < 62.7)$, which is equivalent to $P(Z<-0.0535)$. Using a standard normal table or calculator, $P(Z < - 0.0535)\approx0.4783$.

Step3: Calculate z - score for sample mean

The formula for the z - score of the sample mean $\bar{x}$ is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$. Here, $\bar{x}=62.7$, $\mu = 65.3$, $\sigma=48.6$, and $n = 125$. First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{48.6}{\sqrt{125}}\approx\frac{48.6}{11.1803}\approx4.346$. Then, $z=\frac{62.7 - 65.3}{4.346}=\frac{-2.6}{4.346}\approx - 0.5982$.

Step4: Find the probability for sample mean

We want $P(\bar{X}<62.7)$, which is equivalent to $P(Z < - 0.5982)$. Using a standard normal table or calculator, $P(Z<-0.5982)\approx0.2743$.

Answer:

$P(X < 62.7)=0.4783$
$P(\bar{X}<62.7)=0.2743$