SOLUTION: please help me solve this problem In a test on electric bulbs, it was found that the life time of a particular brand was normally distributed with an average life of 2000 hours an

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Question 896989: please help me solve this problem
In a test on electric bulbs, it was found that the life time of a particular brand was normally distributed with an average life of 2000 hours and S.D. of 60 hours. If a firm purchases 2500 bulbs, find the number of bulbs that are likely to last for (i) more than 2100 hours, (ii) less than 1950 hours and (iii) between 1900 and 2100 hours.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
mean is 2000
standard deviation is 60

2500 bulbs are purchased.

find number of bulbs likely to last:

more than 2100 hours:
less than 1950 hours:
between 1900 and 2100 hours:

you need to find the z-score and then find the area under the normal distribution curve that meets the requirements.

the z-scores are calculated from the population as follows:

z-score = (raw score - mean) / standard deviation.

using this formula, you calculate the following z-scores.

2100 hours: z-score = (2100 - 2000) / 60 = 100 / 60 = 1.67
1950 hours: z-score = (1950 - 2000) / 60 = -50 / 60 = -.83
1900 hours: z-score = (1900 - 2000) / 60 = -100 / 60 = -1.67
2100 hours: z-score = (2100 - 2000) / 60 = 100 / 60 = 1.67

now that you have the z-scores, you need to look up the area under the normal distribution curve that is to the left of these z-scores.

the area to the left of these z-scores is the percent of the distribution that has a z-score less than the indicated z-score.

the areas under the normal distribution curve that are to the left of the indicated z-scores are:

2100 hours: area to the left of a z-score of 1.67 = .9525
1950 hours: area fo the left of a z-score of -.83 = .2033
1900 hours: area to the left of a z-score of -1.67 = .0475

from these z-scores you can find the information that you need.

the probability of the bulb lasting more than 2100 = 1 - .9525 = .0475
the probability of the bulb lasting less than 1950 hours = .2033
the probability of the bulb lasting between 1900 hours and 2100 hours = .9525 - .0475 = .905

how were these probabilities calculated?

raw score of 2100 gives a z-score of 1.67 and the area to the left of that z-score is .9525. this means that the area to the right of that z-score is equal to 1 - .9525 which is equal to .0475. the area to the left of the z-score represents the probability that the score will be less than 2100. the area to the right ot the z-score represents the probability that the score will be greater than 2100.

raw score of 1950 gives a z-score of .83 and the area to the left of that z-score is .2033. that represents the probability that the raw score will be less than 1950. no further adjustment is necessary.

raw score of 1900 gives a z-score of -1.67 and the area to the left of that z-score is .0475. raw score of 2100 gives a z-score of 1.67 and the area to the left of that z-score is .9525. to get the area of the distribution between 1900 and 2100, you have to subtract .0475 from .9525 to get .905.

the following pictures of the distribution curve will show you visually what is happening.

that z-score calculator can be found at the following link:

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

the first picture is the area under the distribution curve to the left of a z-score of 1.67

th second picture is the area under the distribution curve to the left of a z-score of -1.67

the third picture is the area under the distribution curve to the left of a z-score of -.83

the fourth picture is the area under the distribution curve in between a z-score of -1.67 and 1.67.

remember that this area is the difference between the the area to the left of the z-score of -1.67 and the area to the left ot a z-score of 1.67 which is the area to the left of a raw score of 1900 and the raw score of 2100.

now that you have the probabilities, you can calculate what the problem is asking.

out of 2500 bulbs:

the number of bulbs that are expected to last more than 2100 hours is equal to .0475 * 2500 = 118.75.

.0475 represented the probability that the bulb will have a z-score greater than 1.67 and a z-score of 1.67 represented a raw score of 2100.

the number of bulbs that are expected to last less than 1950 hours is equal to .2033 * 2500 = 508.25

.2033 represented the probability that the bulb will have a z-score less than .83 and a z-score of .83 represented a raw score of 1950.

the number of bulbs that are expected to last between 1900 and 2100 hours is equal to .905 * 2500 = 2262.5

.905 represented the probability that the bulb will have a z-score between -1.67 and 1.67 and a z-score of -1.67 represented 1900 bulbs and a z-score of 1.67 represented 2100 bulbs, so .905 represented the probability that the bulbs would live between 1900 and 2100 hours.

here's the pictures: