What is the probability that the average lifetime of 25 bulbs would be no more than 1550 given mean of 1500 and standard deviation of 200 hours.

I know that it is asking P(x less than or equal to 1550), x is 25, u equal 1500 and sd is 200. I'm having problems getting to the final answer.
Thank YOu!

poke around here, and I think all will become clear:

http://davidmlane.com/hyperstat/z_table.html

To find the probability that the average lifetime of 25 bulbs would be no more than 1550, given a mean of 1500 and a standard deviation of 200 hours, we can use the Central Limit Theorem.

The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution.

To apply the Central Limit Theorem, we first calculate the standard deviation of the sampling distribution of the mean, also known as the standard error (SE), using the formula:

SE = standard deviation / square root of sample size

In this case, the standard deviation is 200 hours, and the sample size is 25. So we have:

SE = 200 / sqrt(25) = 200 / 5 = 40

Next, we calculate the z-score, which represents the number of standard deviations away from the mean. The formula for z-score is:

z = (x - mean) / SE

In this case, x is 1550, mean is 1500, and SE is 40. Thus:

z = (1550 - 1500) / 40 = 50 / 40 = 1.25

Now, we can look up the probability corresponding to the z-score of 1.25 in the standard normal distribution table. The table will give us the probability of a z-score being less than or equal to 1.25.

Using the standard normal distribution table, the probability corresponding to a z-score of 1.25 is approximately 0.8944.

Therefore, the probability that the average lifetime of 25 bulbs would be no more than 1550 is approximately 0.8944, or 89.44%.