Continuous probability distributions

Question 1

difference between discrete and continuous data

Answer 1

—Remember: discrete data has a countable number of possible values –>discrete probability distributions can be put in tables
—Continuous data have an infinite number of possible values –>we use a smooth function, f(x) to describe the probabilities

Question 2

what is a good way to represent continuous data?

Answer 2

via histograms

Question 3

how do observations influence the shape of the histogram?

Answer 3

more observations result to a smoother histogram

Question 4

how to represent a continuous random variable?

Answer 4

continuous random variable is a function such that the probability that the variable lies in an interval (a, b) is the area under the curve from a to b

Question 5

what are features of a probability density function?

Answer 5

—f(x) must satisfy the following:

1. f(x)≥0 for all x, that is, it must be non-negative.
2. The total area underneath the curve representing f(x) = 1.

Question 6

what do areas on the histogram represent?

Answer 6

—In each case above, the % areas of the histogram boxes (that is, the area of a box as a % of the total area of all the boxes) are providing estimates of the probabilities of intervals.

Question 7

what are possible shapes of f(x)?

Answer 7

Question 8

how to evaluate area under curve?

Answer 8

Question 9

for continuous probability density function, when x takes on a specific value eg. x=2, what is the area?

Answer 9

area = 0

Question 10

describe the mean and variance of a continuous random variable

Answer 10

The mean measures the location of the distribution, the variance measures the spread of the distribution.

Question 11

Find P(X>0.5)

Answer 11

¼

Question 12

Find P(X<0.75)

Answer 12

—1-P(X>0.75)=1-(½*¼*½)=15/16

Question 13

what is uniform distribution?

Answer 13

—Special sort of distribution for continuous data.
—Described by the function
f(x) = 1/b-a a ≤ x ≤ b

Question 14

what is expectation and variance of uniform distribution?

Answer 14

Question 15

—The length of time patients wait to see a doctor is uniformly distributed between 40 minutes and 3 hours.
—Let X be the waiting time in minutes.

Answer 15

as f(x) = 1/b-a

f(x) = 1/180-40

f(x)=1/140

40 less than or equal to X 180 equal to or greater than

Question 16

—The length of time patients wait to see a doctor is uniformly distributed between 40 minutes and 3 hours.
—Let X be the waiting time in minutes.
Find the probability of waiting between 50 minutes and 2 hours.

Answer 16

—P(50≤x≤120) = (120-50) x 1/140 = 0.5

Question 17

—The length of time patients wait to see a doctor is uniformly distributed between 40 minutes and 3 hours.
—Let X be the waiting time in minutes.

Find the mean and variance of the distribution of waiting times

Answer 17

Question 18

—The length of time patients wait to see a doctor is uniformly distributed between 40 minutes and 3 hours.
—Let X be the waiting time in minutes.
Find the probability of having to wait exactly one hour.

Answer 18

—P(X=60) = 0

Question 19

what is another special sort of distribution for continuous data?

Answer 19

The Normal Distribution

—The general form of the pdf (probability density function) is given by:

Question 20

describe normal distribution

Answer 20

—Bell-shaped, symmetric about µ, reaches highest point at x=µ, tends to zero as x→±∞.

Question 21

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About the Normal Distribution<!--EndFragment-->

Answer 21

1. E(X) = µ; V(X) = σ².
2. Area under curve = 1
3. Different means – shift curve up and down x-axis
4. Different variances – curve becomes more peaked or more squashed
5. Shorthand notation: X~N(µ, σ²).

Question 22

what do different means on a normal distribution look like?

Answer 22

Question 23

what do different variances on normal distribution look like?

Answer 23

Question 24

what is the standard normal distributoin?

Answer 24

when μ=0, σ² =1

Question 25

P(Z\<-1.08)

Answer 25

3) P(Z\<-1.08) = P(Z\>1.08) by symmetry = 0.1401 (from (2)) OR P(Z\<-1.08) = 0.1401 from tables directly.

Question 26

P(-1.51

Answer 26

= P(Z = 0.8599 – 0.0655 = 0.7944

Question 27

what do you call a random variable from a standard normal distribution?

Answer 27

—Call a r.v. from this a Standard Normal r.v., use notation Z~N(0,1)

Question 28

how do you convert any Normal random variable to a Standard Normal Random Variable?

Answer 28

—If X~N(μ,σ²), then use the linear transformation below: the standardised value is known as the Z score —So, for ANY random variable that comes from a normal distribution, if we subtract the mean and divide by the standard deviation, we get a r.v.~N(0,1).

Question 29

Find P(Z\>1.08)

Answer 29

= 1 – P(Z\<1.08) = 1 – 0.8599 (from tables) = 0.1401

Question 30

—Suppose X~N(50, 100²). Find P(X\<200)

Answer 30

Question 31

—The length of metallic strips produced by a machine are normally distributed with a mean of 100cm and a variance of 2.25cm². Only strips that are between 98 and 103cm are acceptable. What proportion of strips are acceptable?

Answer 31

Question 32

—A battery lasts an average of 3 years with a standard deviation of 0.5 years. Assuming that battery lives are normally distributed, find the probability that a given battery will last less than 2.3 years.

Answer 32

Question 33

—Salaries of workers in a factory are normally distributed with mean $48,000 and standard deviation $3,500. What is the minimum salary of the top 20% of workers?

Answer 33

—From tables, P(Z So, P(Z\> (a-48000/3500) ) = 0.2 P(Z\<0.84) = 0.8 (need to find area below) P(Z\>0.84) = 0.2 (Area above) Therefore a-48000/3500 = 0.84 a= 50940

Question 34

features of continuous probability distributions

Answer 34

1. f(x)≥0 for all x 2. Area under curve = 1

Question 35

describe symmetry found in Z tables

Answer 35

—Symmetry → P(Z\<-a) = P(Z\>a)

Question 36

1. The time of failure for a continuous operation monitoring device of air quality has a uniform distribution over a 24 hour day If the device has a self-checking computer chip that determines whether the device is operational every hour on the hour, what is the probability that it will take at least 40 minutes to detect that a failure has occurred?

Answer 36

``` P(Failure takes longer than 40 minutes to detect) = P(failure occurs between 12-12.20am; or 1-1.20am; or 2-2.20am …) Hence P(failure takes longer than 40 minutes to detect = 24 \* 1/24 \* 20/60 = 1/3 ```

Question 37

1. The time of failure for a continuous operation monitoring device of air quality has a uniform distribution over a 24 hour day If the device has a self-checking computer chip that determines whether the device is operational every hour on the hour, what is the probability that a failure will be detected within 10 minutes of its occurrence?

Answer 37

P(failure detected within 10 minutes) = P(failure occurs between 12.50-1am; or 1.50-2am; or 2.50-3am ……) There are 24 windows each 10 minutes wide during which the failure could occur to be detected within 10 minutes. Hence P(failure detected within 10 minutes) = 24 \* 1/24 \* 10/60 = 1/6 d. If the devi

Question 38

The time of failure for a continuous operation monitoring device of air quality has a uniform distribution over a 24 hour day. a. If a failure occurs on a day when it is daylight between 5.55am and 7.38pm, what is the probability that the failure will occur during daylight hours?

Answer 38

Let X be the time of failure, past midnight, in hours. X has a uniform distribution, with positive probability from a lower bound of 0 to an upper bound of 24. **f(x) = 1/24, 0
P(failure occurs in daylight) = P(failure is between 5.55am and 
7.38am) 
This probability is calculated by finding the area of the rectangle 
with height 1/24 and base length (5.55am to 7.38pm) = 13 hours 
and 43 minutes = 13 43/60 hours. 
Hence P(failure in daylight) = 1/24 * (13 43/60) = 823/1440 **