Problem
Given keyword: “let x represent the number of minutes”. Interpret this as a quantitative random variable for time.
Let \(X\) represent the number of minutes required for a student to finish a short statistics quiz. Assume completion time is normally distributed with mean \(\mu=18\) minutes and standard deviation \(\sigma=3\) minutes.
- (a) Find \(P(X \le 20)\).
- (b) Find \(P(15 < X < 22)\).
- (c) Find the time \(t\) such that only \(10\%\) of students take longer than \(t\).
Key idea: standardizing a normal distribution
Any value \(x\) (in minutes) can be converted to a standard normal value \(z\) using:
\[ z=\frac{x-\mu}{\sigma} \]
Then probabilities about \(X\) become probabilities about \(Z\), where \(Z\) follows the standard normal distribution.
Solution
(a) Compute \(P(X \le 20)\)
Standardize \(x=20\): \[ z=\frac{20-18}{3}=\frac{2}{3}=0.6667 \]
So \(P(X \le 20)=P(Z \le 0.6667)\). Using the standard normal cumulative probability, \[ P(Z \le 0.6667)\approx 0.7475 \]
Answer (a): \(P(X \le 20)\approx 0.7475\).
(b) Compute \(P(15 < X < 22)\)
Standardize both bounds: \[ z_{15}=\frac{15-18}{3}=\frac{-3}{3}=-1 \] \[ z_{22}=\frac{22-18}{3}=\frac{4}{3}=1.3333 \]
Convert to a standard normal difference:
\[
P(15
Using standard normal cumulative probabilities:
\[
P(Z<1.3333)\approx 0.9088,\quad P(Z<-1)\approx 0.1587
\]
\[
P(15 Answer (b): \(P(15
The condition \(P(X>t)=0.10\) is equivalent to \(P(X\le t)=0.90\).
Let \(z_{0.90}\) be the \(90\)th percentile of the standard normal distribution:
\[
P(Z\le z_{0.90})=0.90,\quad z_{0.90}\approx 1.2816
\]
Convert back to minutes:
\[
t=\mu+z_{0.90}\cdot\sigma=18+1.2816\cdot 3=21.8448
\]
Answer (c): \(t\approx 21.84\) minutes (about \(21.85\) minutes).(c) Find \(t\) such that \(P(X>t)=0.10\)
Summary table
Part
Event in minutes
Standardized form
Approximate value
(a)
\(X \le 20\)
\(Z \le 0.6667\)
\(0.7475\)
(b)
\(15 < X < 22\)
\(-1 < Z < 1.3333\)
\(0.7501\)
(c)
\(P(X>t)=0.10\)
\(t=18+z_{0.90}\cdot 3\)
\(t\approx 21.85\)
Interpretation