–/3 points Notes Question: TanColM7 8.6.003.
TKK Products manufactures electric light bulbs in the 50-, 60-, 75-, and 100-watt range. Laboratory tests show that the lives of these light bulbs are normally distributed with a mean of 774 hr and standard deviation of 68 hr. Find the probability that a TKK light bulb selected at random will burn for the following times. Round your standard normal variable to two decimal places before using the table of values. (Give your answers to four decimal places.)
(a) less than 842 hr
(b) between 842 and 910 hr
(c) more than 638 hr
To find the probability for each scenario, we need to convert the given values to z-scores and then use the standard normal distribution table to determine the probabilities.
The formula to calculate the z-score is:
z = (x - mean) / standard deviation
First, let's calculate the z-scores for each scenario:
(a) To find the probability of the light bulb burning for less than 842 hours:
z = (842 - 774) / 68
z ≈ 1.00
(b) To find the probability of the light bulb burning between 842 and 910 hours:
z1 = (842 - 774) / 68 ≈ 1.00
z2 = (910 - 774) / 68 ≈ 2.00
(c) To find the probability of the light bulb burning for more than 638 hours:
z = (638 - 774) / 68
z ≈ -2.00
Now, we will use the standard normal distribution table to find the probabilities associated with these z-scores.
(a) The probability of the light bulb burning for less than 842 hours can be found by looking up the z-score of 1.00 in the table. Using the table, we find that the probability is approximately 0.8413.
(b) The probability of the light bulb burning between 842 and 910 hours is the difference between the probabilities associated with the z-scores of 2.00 and 1.00. Using the table, we find that the probability for z = 2.00 is approximately 0.9772, and the probability for z = 1.00 is approximately 0.8413. Therefore, the probability is approximately 0.9772 - 0.8413 = 0.1359.
(c) The probability of the light bulb burning for more than 638 hours can be found by looking up the z-score of -2.00 in the table and subtracting the result from 1 (total probability). Using the table, we find that the probability for z = -2.00 is approximately 0.0228. Therefore, the probability is approximately 1 - 0.0228 = 0.9772.
In summary, the probabilities are:
(a) P(X < 842) ≈ 0.8413
(b) P(842 < X < 910) ≈ 0.1359
(c) P(X > 638) ≈ 0.9772