The waiting times for service calls made to a computer techn

The waiting times for service calls made to a computer technical support line are known to be 15 min. on average, with a standard deviation of 2 min.

a. assuming nothing about the data, at least 24/25 proportion of the waiting times will fall between ____ and ____ min

b. according to chebyshev\'s rule at least _____ percent of the waiting times will fall between 11.5 and 18.5

c. what can you say about the % of the waiting times that will take more than 20 min

d assuming the data comes from a mound shaped distribution approx. what percent of the waiting times will be between 11-17 min

e. according to the empirical rule, approx 16% of the waiting times will be greater than ___ min

f. given that the data comes from a mound-shaped distribution, approxmiately __ proportion of the data will fall between 9 and 21 min

We got M = 15

s = 2

Answer to part a)

24/25 proprotion = 0.96 , thus we need to find out the values which include 0.96 portion of the data within

this means outside this range under both the tails the total area would be 0.04 , does each tail has 0.02 area

we need to find the Z value corresponding to left area = 0.02 from the Z table

We get Z = -2.05 and Z = +2.05

Thus now we can get the lower and the upper limits of this range

The Z formula is :

Z = x - M / s

-2.05 = x - 15 / 2

-4.10 = x - 15

x = 15 - 4.10

x = 10.90

2.05 = x - 15 /2

4.10 = x - 15

x = 15 + 4.10

x = 19.10

Answer to part b)

15 -2*2 = 11 and 15 +2*2 = 19

Thus atleast 75% of the waiting time lies between 2 standard deviations

Answer to part c)

The percent of time more than 20 mins will be as follows

P(x > 20) = 1 - P(x < 20)

P(x < 20) = P(z < 20-15 /2)

P(x < 20) = P(z < 2.5)

From the Z table we get to know that

P(x < 20) = P(z < 2.5) = 0.9938

Thus P(x > 20) = 1 - 0.9938

P(x > 20) = 0.0062

In terms of percent it would be : 0.0062 *100 = 0.62%

Answer to part d)

We need to find P(11 < x < 17) = P(x < 17) - P(x < 11)

P(x < 17) = P(z < 17 -15 /2) = P( z < 1 ) = 0.8413

P(x < 11) = P(z < 11 - 15 / 2) = P(z < -2) = 0.0228

Thus P(11 < x < 17) = 0.8413 - 0.0228 = 0.8185