Question

1. the mean life of a forever bright lightbulb is found to be 700 hours with a standard deviation of 140 hours. if meaghan, an employee at an amusement park, buys and installs 10,000 such lightbulbs, find the number that can be expected to last:

a) less than 800 hours. (3 marks)
μ = 700hrs
σ = 140hrs
x = 10,000

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 800\), \(\mu=700\), and \(\sigma = 140\).

$$z=\frac{800 - 700}{140}=\frac{100}{140}\approx0.71$$

Step2: Find the probability using the standard normal distribution table

Looking up the value of \(z = 0.71\) in the standard normal distribution table. The probability \(P(Z<0.71)\) is \(0.7611\) (from the standard normal table: the value corresponding to \(z = 0.71\) is \(0.7611\)).

Step3: Calculate the number of light - bulbs

If \(n = 10000\) light - bulbs are installed, and the probability that a light - bulb has a life less than \(800\) hours is \(P(X < 800)=P(Z<0.71) = 0.7611\). Then the number of light - bulbs \(N\) that can be expected to last less than \(800\) hours is \(N=n\times P(X < 800)\).

$$N=10000\times0.7611 = 7611$$

Answer:

\(7611\)