Question 854240: A statistical analysis of a telephone calls made by a company indicates that the length of normally distributed, with mean( u=260 seconds) and variance =40 seconds.
(a) what is the probability that a call lasted between 190 and 340 seconds?
(b) what is the length of a call if only 1% of calls are shorter ?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean = 260
variance = 40
standard deviation = square root of variance = square root (40) = 6.324555 rounded to 6 decimal places.
(a) what is the probability that a call lasted between 190 and 340 seconds?
(b) what is the length of a call if only 1% of calls are shorter ?
z-score = (x-m)/s
x is the score
m is the mean
s is the standard deviation
you are looking at 2 z-scores and the probability that the call will be between them.
x1 = 190 seconds
x2 = 340 seconds
z1 = (x1 - m) / s
x1 = 190
m = 260
s = 6.324555
z1 = (190 - 260) / 6.324555
this makes z1 = -11.06797237
that looks awfully big z.
if you meant standard deviation = 40 then the results would look more normal.
i'll assume you meant standard deviation = 40 and work the problem that way because variance equal 40 just doesn't make much sense.
standard deviation is equal to the square root of the variance so variance of 40 gives us a standard deviation that's just too small for this type of problem.
i suspect standard deviation is 40 would be more in line with what it should be rather than variance = 40.
assuming that the standard deviation is 40, the results would be as follows:
z1 = (190 - 260) / 40 = -1.75
z2 = (340 - 260) / 40 = 2
you would look up the z-score of -1.75 and the z-score of 2 and you would then subtract the area to the left of these z-scores to get the area in between and you would get:
p(z2-z1) = .9772 - .0401 = .9371
there is a 93.71% probability that a the z-score you get will be between a z-score of -1.75 and 2.
put into words of real scores, there is a 93.71% probability that the duration of the call will be between 190 and 340 seconds.
the table i used is at the following link:
http://lilt.ilstu.edu/dasacke/eco148/ztable.htm
if you want to find the duration of the call where only 1% of the calls are shorter, you would use this table again and find the z-score that gives you an area to the left of it that is closest to .01
in that table, i found that an area to the left of the z-score that gave an area to the left of it closest to .01 was equal to -2.33.
since z-score = (x-m)/s, it follows that x = s * z-score + m.
assuming s = 40 and m = 260 and z = -2.33, this formula yields:
x = 40 * -2.33 + 260 which is equal to 166.8
we can back track with that value for x and see if we did it right.
z = (x - m) / s which becomes z = (166.8 - 260) / 40 which becomes z = -2.33.
we then look up in the table for the area to the left of that and we get .0099, so we confirmed that we did it correctly.
that'a about the best that we can do given the accuracy of the tables and it's close enough for the type of problem that we're dealing with.
i used my TI-84 calculator to find the z-score for an area to the left of it of .01 and i got z = -2.326347877.
round that to 2 decimal places and you get -2.33 so we did pretty good.
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