Business Statistics Exam Study Guidance

Q2. What StatCrunch calculator instruction is needed to determine the lifetime that should be advertised for flood lamps, given a mean of 3750 hours, standard deviation of 300 hours, and a requirement that no more than 2% burn out before the advertised lifetime?

Background

Topic: Normal Distribution & Probability Calculation

This question tests your ability to use the normal distribution to determine a cutoff value (advertised lifetime) so that only a small percentage (2%) of lamps fail before that time. This is a common application of probability in quality control and product guarantees.

Key Terms and Formulas

• Normal Distribution: A continuous probability distribution characterized by a symmetric, bell-shaped curve.

• Mean ($\mu$): The average value, here 3750 hours.

• Standard Deviation ($\sigma$): The spread of the distribution, here 300 hours.

• Probability ($P$): The likelihood that a lamp burns out before a certain time.

• Z-score formula: $Z = \frac{X - \mu}{\sigma}$

Step-by-Step Guidance

1. Identify the requirement: No more than 2% of lamps should burn out before the advertised lifetime. This means you want the advertised lifetime to be at the 2nd percentile of the normal distribution.

2. Set up the probability statement: $P(X \leq \text{Advertised Lifetime}) = 0.02$

3. Use the normal calculator in StatCrunch: Enter the mean (3750) and standard deviation (300), and set the probability to 0.02 for the left tail.

4. Find the value of $X$ such that $P(X \leq X) = 0.02$. This is the cutoff for the advertised lifetime.

Try solving on your own before revealing the answer!

Final Answer: StatCrunch instruction is P(X ≤ 3133.8753) = 0.02

This means the manufacturer should advertise a lifetime of approximately 3133.88 hours to ensure no more than 2% of lamps burn out before this time.