Question

1. cost of personal computers the average price of a personal computer (pc) is $949. if the computer prices are approximately normally distributed and $\sigma = \\$100$, what is the probability that a randomly selected pc costs more than $\\$1200$? the least expensive 10% of personal computers cost less than what amount?

Explanation:

Part 1: Probability a PC costs more than $1200

We are given a normal distribution with mean \(\mu = 949\) and standard deviation \(\sigma = 100\). We need to find \(P(X>1200)\).

Step 1: Calculate the z - score

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\).
For \(x = 1200\), we have:
\(z=\frac{1200 - 949}{100}=\frac{251}{100}=2.51\)

Step 2: Find the probability

We want \(P(X > 1200)\), which is equivalent to \(P(Z>2.51)\) (where \(Z\) is the standard normal variable).
Since the total area under the standard normal curve is 1, \(P(Z > 2.51)=1 - P(Z\leqslant2.51)\).
Looking up the value of \(P(Z\leqslant2.51)\) in the standard normal table (z - table), we find that \(P(Z\leqslant2.51) = 0.9940\).
So, \(P(Z > 2.51)=1 - 0.9940=0.0060\)

Part 2: The least expensive 10% of PCs

We need to find the value \(x\) such that \(P(X < x)=0.10\).

Step 1: Find the z - score corresponding to the 10th percentile

Looking up in the standard normal table, the z - score \(z\) such that \(P(Z < z)=0.10\) is approximately \(z=- 1.28\) (since the area to the left of \(z=-1.28\) is about 0.10).

Step 2: Use the z - score formula to find \(x\)

The z - score formula is \(z=\frac{x-\mu}{\sigma}\). We know \(z=-1.28\), \(\mu = 949\) and \(\sigma = 100\).
Rearranging the formula for \(x\): \(x=\mu+z\sigma\)
Substitute the values: \(x = 949+(-1.28)\times100=949 - 128 = 821\)

Answer:

s:

• The probability that a randomly selected PC costs more than $1200 is \(\boldsymbol{0.0060}\).
• The least expensive 10% of personal computers cost less than \(\boldsymbol{\$821}\).