The distribution of lifetimes for a particular brand of lightbulbs has a mean of 110 hours and a standard deviation of 20 hours. Suppose a random sample of 35 lightbulbs is taken for observation.
What is the probability that the sample mean lifetime will be between 105 and 115 hours?
0.4306
0.3026
0.1974
0.7372
0.8612
What value does the sample mean exceed with 95% probability (Hint: This is the 5th percentile for the sample mean distribution)?
77.10
106.79
104.44
116.63
115.56
For an individual bulb, the probability of a life between 105 and 115 hours is 0.1974. I used a normal probability calculator to get that.
For 35 bulbs, the mean of the average remains 110 but the standard deviation is 20/sqrt35 = 3.38. The probabilty of the mean being between 105 and 115 is then 0.861.
For your second question, I get 104.4
To find the probability that the sample mean lifetime is between 105 and 115 hours, we can use the Central Limit Theorem and approximate the distribution of sample means to a normal distribution.
The mean of the sample mean distribution is equal to the population mean, which is 110 hours. The standard deviation of the sample mean distribution is calculated by dividing the population standard deviation by the square root of the sample size, which is 20/sqrt(35) ≈ 3.37.
To find the probability, we need to calculate the z-scores for the lower and upper limits of the desired range and use the standard normal distribution table.
The z-score for 105 hours is (105 - 110) / 3.37 ≈ -1.48
The z-score for 115 hours is (115 - 110) / 3.37 ≈ 1.48
Looking up the z-scores in the standard normal distribution table, we find that the probability associated with a z-score of -1.48 is 0.0694, and the probability associated with a z-score of 1.48 is 0.9306.
To find the probability between 105 and 115 hours, we subtract the probability associated with the lower z-score from the probability associated with the upper z-score:
0.9306 - 0.0694 = 0.8612
Therefore, the probability that the sample mean lifetime will be between 105 and 115 hours is 0.8612. So, the correct option is 0.8612.
Now, to determine the value that the sample mean exceeds with 95% probability, we need to find the z-score associated with a cumulative probability of 0.05 (since it's the 5th percentile).
Using the standard normal distribution table, we find that the z-score associated with a cumulative probability of 0.05 is approximately -1.645.
To find the value of the sample mean, we use the formula: sample mean = population mean + (z-score * standard deviation / sqrt(sample size)).
sample mean = 110 + (-1.645 * 20 / sqrt(35)) ≈ 104.44
Therefore, the value that the sample mean exceeds with 95% probability is approximately 104.44. So, the correct option is 104.44.