Question

a population of rug prices has a normal distribution with $mu = 94.7$ and $sigma = 45.2$. you intend to draw a random sample of size $n = 217$. please show your answers as numbers accurate to 4 decimal places. find the probability that a single randomly selected rug is greater than 101.1. $p(x>101.1)=$ find the probability that a sample of size $n = 217$ rugs are randomly selected with a mean greater than 101.1. $p(overline{x}>101.1)=$ question help: video message instructor post to forum submit question

Explanation:

Step1: Calculate z - score for single - value

The formula for the z - score of a single value $X$ in a normal distribution is $z=\frac{X-\mu}{\sigma}$. Here, $\mu = 94.7$, $\sigma = 45.2$, and $X = 101.1$. So, $z=\frac{101.1 - 94.7}{45.2}=\frac{6.4}{45.2}\approx0.1416$.

Step2: Find $P(X>101.1)$

We know that $P(X > 101.1)=1 - P(X\leq101.1)$. Looking up the z - score of $0.1416$ in the standard normal distribution table, $P(Z\leq0.1416)\approx0.5562$. So, $P(X > 101.1)=1 - 0.5562 = 0.4438$.

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, $\mu = 94.7$, $\sigma = 45.2$, $n = 217$, and $\bar{X}=101.1$. First, calculate $\frac{\sigma}{\sqrt{n}}=\frac{45.2}{\sqrt{217}}\approx3.0877$. Then, $z=\frac{101.1 - 94.7}{3.0877}=\frac{6.4}{3.0877}\approx2.0727$.

Step4: Find $P(\bar{X}>101.1)$

We know that $P(\bar{X}>101.1)=1 - P(\bar{X}\leq101.1)$. Looking up the z - score of $2.0727$ in the standard normal distribution table, $P(Z\leq2.0727)\approx0.9808$. So, $P(\bar{X}>101.1)=1 - 0.9808 = 0.0192$.

Answer:

$P(X > 101.1)=0.4438$
$P(\bar{X}>101.1)=0.0192$