Suppose that the certain lifetimes of a certain light bulb are normally distributed with μ=1500 hours and σ=200. Find the probability that a light bulb will burn out in less than 1200 hours.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.
Z=(X<1.5)=0.4332+0.5=0.9332
To find the probability that a light bulb will burn out in less than 1200 hours, we need to calculate the z-score and then use the standard normal distribution.
The z-score formula is given as:
z = (X - μ) / σ
where X is the value we want to find the probability for (1200 hours in this case), μ is the mean of the distribution (1500 hours), and σ is the standard deviation of the distribution (200 hours).
Substituting the given values into the formula, we get:
z = (1200 - 1500) / 200
z = -300 / 200
z = -1.5
Next, we need to find the probability corresponding to this z-score from the standard normal distribution table or calculator. The table or calculator will give us the probability associated with the area to the left of the z-score.
Looking up the z-score of -1.5 in the standard normal distribution table, we find that the probability is approximately 0.0668.
Therefore, the probability that a light bulb will burn out in less than 1200 hours is approximately 0.0668 or 6.68%.