Question

question 2
a company owned vehicle is used for employee trips. the vehicle is used an average of 10 days per
month, with a standard deviation of 2 days. data is normally distributed. four decimal places
find the probability that the car is used more than 15 days per month
find the probability that the car is used between 12 and 15 days per month
find the probability that the car is used between 6 and 15 days per month
find the probability that the care is used more than 5 days per month

Explanation:

Step1: Calculate z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 10\) (mean), \(\sigma=2\) (standard deviation)

Part 1: Probability that the car is used more than 15 days per month

Step2: Calculate z - score for \(x = 15\)

\(z=\frac{15 - 10}{2}=\frac{5}{2}=2.5\)

Step3: Find probability using standard normal table

\(P(X>15)=1 - P(X\leq15)\)
From the standard normal table, \(P(Z\leq2.5)=0.9938\)
\(P(X > 15)=1-0.9938 = 0.0062\)

Part 2: Probability that the car is used between 12 and 15 days per month

Step2: Calculate z - scores

For \(x = 12\): \(z_1=\frac{12 - 10}{2}=1\)
For \(x = 15\): \(z_2 = 2.5\) (calculated above)

Step3: Find probabilities using standard normal table

\(P(12<X<15)=P(Z < 2.5)-P(Z < 1)\)
From the standard normal table, \(P(Z<1)=0.8413\), \(P(Z < 2.5)=0.9938\)
\(P(12<X<15)=0.9938 - 0.8413=0.1525\)

Part 3: Probability that the car is used between 6 and 15 days per month

Step2: Calculate z - scores

For \(x = 6\): \(z_1=\frac{6 - 10}{2}=\frac{- 4}{2}=-2\)
For \(x = 15\): \(z_2 = 2.5\) (calculated above)

Step3: Find probabilities using standard normal table

\(P(6<X<15)=P(Z < 2.5)-P(Z<-2)\)
From the standard normal table, \(P(Z < - 2)=0.0228\), \(P(Z < 2.5)=0.9938\)
\(P(6<X<15)=0.9938-0.0228 = 0.9710\)

Part 4: Probability that the car is used more than 5 days per month

Step2: Calculate z - score

For \(x = 5\): \(z=\frac{5 - 10}{2}=\frac{-5}{2}=-2.5\)

Step3: Find probability using standard normal table

\(P(X>5)=1 - P(X\leq5)\)
From the standard normal table, \(P(Z\leq - 2.5)=0.0062\)
\(P(X>5)=1 - 0.0062=0.9938\)

Answer:

1. \(0.0062\)
2. \(0.1525\)
3. \(0.9710\)
4. \(0.9938\)