A computer instructor found out that on average, his students complete their modules in 120 hours with a standard deviation of 10 hours. Assuming that his data are normally distributed, what is the probability that a randomly selected student will complete his module in 100 to 140 hours? Write your answer correct to four decimal places, as in 0.1234.
The correct answer and explanation is:
To solve this problem, we will use the normal distribution and Z-scores to calculate the probability that a randomly selected student will complete their module in 100 to 140 hours.
Step 1: Identify the Mean and Standard Deviation
- Mean (μ) = 120 hours
- Standard deviation (σ) = 10 hours
Step 2: Convert the Hours to Z-Scores
We need to convert the values of 100 and 140 hours to Z-scores using the following formula: Z=X−μσZ = \frac{X – \mu}{\sigma}
Where:
- XX is the value (either 100 or 140 hours)
- μ\mu is the mean (120 hours)
- σ\sigma is the standard deviation (10 hours)
Z-score for 100 hours:
Z=100−12010=−2010=−2Z = \frac{100 – 120}{10} = \frac{-20}{10} = -2
Z-score for 140 hours:
Z=140−12010=2010=2Z = \frac{140 – 120}{10} = \frac{20}{10} = 2
Step 3: Look Up the Z-Scores in the Standard Normal Table
We now look up the Z-scores in the standard normal distribution table or use a calculator.
- The probability corresponding to a Z-score of -2 is approximately 0.0228.
- The probability corresponding to a Z-score of 2 is approximately 0.9772.
Step 4: Find the Probability Between 100 and 140 Hours
To find the probability that a student will complete their module between 100 and 140 hours, subtract the probability corresponding to Z = -2 from the probability corresponding to Z = 2: P(100≤X≤140)=P(Z=2)−P(Z=−2)=0.9772−0.0228=0.9544P(100 \leq X \leq 140) = P(Z = 2) – P(Z = -2) = 0.9772 – 0.0228 = 0.9544
Final Answer:
The probability that a randomly selected student will complete their module in 100 to 140 hours is 0.9544.