8.1(A): Continuous Uniform Distribution
Explain the three different Unimodal Distribution types?
Symmetric, Positively Skewed, Negatively Skewed
Which one of the following is the correct formula?
a) 1+e^kx
b) 1-e^kx
c) 1+e^-kx
d) 1-e^-kx
D) 1-e^-kx
What is the mean, medium and mode of the following set?
3, 6, 1, 5, 8
Mean: 3 + 6 + 1 + 5 + 8 / 5 = 4.6
Medium: 1, 3, 5, 6, 8 = 5 splits the set
Mode: No mode: All values have equal frequency
True or False: Probabilities can be computed by finding the area under the curve within the appropriate interval
True
At the Rogers call centre, the average time between phone calls to its tech support line is 2 min and it is exponentially distributed.
(a) Determine the probability of a sales representative waiting 2 min or less before a call comes.
(b) What is the prob of waiting at most 30 sec between calls?
a)P(X<x) = 1-e^-0.5(2) = 0.6321
63% wait time for waiting 2 mins or less
b) P(X<x) = 1-e^-0.5(1/2) = 0.2212
22% probability of waiting at most 30 sec
Complete the chart below:
Mean : Standard Deviation : Probability
a) 12 3 P(X<9) =
b) 30 5 P(X<25)=
c) 5 2.2 P(X>6) =
a)Z score = 9-12/3 = -1 =0.1587 or 0.16
b)Z score = 25-30/5 = -1= 0.1587 or 0.16
c)Z score = 6-5/2.2 = 0.4545 = 0.6736
1-0.6736 = 0.3264
What is the difference between Bimodal and Unimodal Distributions?
The Unimodal distribution only has a single “hump” and is used for a set of discrete values, while the bimodal distribution has two “humps” and is used when a population consists of two groups with different attributes.
(Ex. The distribution of adult shoe sizes is bimodal because men tend to have larger feet than women do.)
At the Rogers call centre, the average time between phone calls to its tech support line is 2 min and it is exponentially distributed.
Determine the probability of a technician waiting between 2 to 3 minutes before a call comes.
K= 1/mean = ½ or 0.5
P(2<x<3) = P(x<3) – P(x<2)
= [1-e^-0.5(3)] – [1-e^-0.5(2)]
= 0.7768 – 0.6321 = 0.14 or 14% probability that wait is between 2 and 3 minutes.
It is found that a certain brand of Samsung Microwaves have a mean lifetime of μ =215 hrs with a standard deviation of σ =12 hrs. If the lifetime of these microwaves is normally distributed, what percentage of the toasters would you expect to:
a) malfunction in less than 200 hours
b) Lifetime of more than 250 hours
a) Z score: 200-215/12 = -1.25 = 0.1056 or Probability there is a malfunction before 200 hours is 10.56%.
b) P(X>250 -215/ 12) = P(X> 2.92) = 1-P(X> 2.92 = 1- 0.9982 = 0.0018 = 18% probability the microwaves lasts more than 250 hours
The drive time between Toronto and North Bay is found to range evenly between 195 and 240 min. What is the prob that the drive will take less than 210 min?
195< x< 240
Area of Rectangle= 1
w= (1/240-195), therefore w= (1/45)
A= (210-195) x (1/45)
= 15 x (1/45)
= ⅓ -> 33.3%
The Lifetime of an iPhone is exponentially distributed, with the manufacturer claiming that it lasts an average of 5 Years before it starts to malfunction
A) What is the probability that it will start to malfunction in less than 90 days
B) What is the probability that it will last longer than 6 years
A) P( X=x) = 1- e^-kx
P(X<90/365)= 1- e^-(1/5)(90/365)= 0.04811 (4.8%, approximately 5%)
B) P( X=x) = 1- e^-kx
P(X<6)= 1- e^-(1/5)(6)= 0.6988 (69.8% or approximately 70%)
Complete the chart below:
Mean : Standard Deviation : Probability
a) 10.5 0.95 P(X<9.1)
b) 10.5 0.95 P(10.1<X<10.9)
a) P(X<9.1) = P(X< 9.1-10.5/0.95)
= P(X < -1.47) = p(X < 0.0708)
= 7.08%
b) P(10.1<X<10.9) = P(10.1-10.5/0.95) < X < P(10.9-10.5/0.95) = P(-0.42 < X < 0.42) = P(0.3372 < X < 0.6628) =0.6628 - 0.3372 = 0.3256 = 32.56 %
The download time from YouTube is found to range evenly between 22 and 39 seconds. What is the probability that the download will take less than 30 seconds?
22 < x < 39
w=1
(39-22) x W= 1
w= (1/39-22) -> (1/17)
A= (30-22)
= 8 x (1/17)
= 0.4705 (47%)
The probability that the download time will take less than 30 seconds is 47%
Arsh has just received admission to the University of Toronto for Rottman Commerce. She has a question about OSAP and calls the office of Financial Services at UofT. When she calls, all the lines are busy, and the automated response puts her on hold and advises her that the waiting time is exponentially distributed with a mean waiting time of 14 minutes. What is the probability that she…
A) Less than 8 Minutes
B) More than 8 Minutes
C) Between 10-20 Minutes
D) Arsh called at 4:15 pm in the afternoon. The office closes at 4:30 pm, and if a caller is still on hold by 4:30, they are automatically disconnected. What is the probability that Arsh will get disconnected, and will be forced to call back tomorrow?
P(X<8)= 1- e^-(1/14)(8)= 0.435 (Approximately 44%)
P(X>8)= 1- P(X<8) ( 1- 0.435= 0.5647 or 56%)
P(10 < X < 25)= P(X < 25) - P(X< 10) (0.7603 - 0.5015= 0.2498 (Approximately 25%))
P(X>15) = 1- e^-(1/14)(15)= 0. 6574 (Aprroximately 66%)
A class of 135 took a final exam in Advanced Functions. The mean score on the exam was 68% with a standard deviation of 8.5%. Determine the percentile rank of each of the following people:
a) Deep who scored 78%
b) Axar, who got 55%
c) Jasdeep, who was happy with his 89%
A) P(X<78) = 78-69/8.5 = 1.17647 = 0.8790
= 87.9% or 88 percentile
B) P(X<55) = 55-68/8.5 = -1.5294 = 0.0643
= 6.43% or 6th percentile
C) P(X<89) = 89-68/8.5 = 2.470 = 0.9932
= 99th percentile