Skip to content

2 Probability: Continuous

2.9 Continuous random variable

a)

The following R commands and results are given: 

pnorm(2)
[1] 0.9772
pnorm(2,1,1)
[1] 0.8413
pnorm(2,1,2)
[1] 0.6915
Specify which distributionsare used and explain the resulting probabili- ties (preferably by a sketch).

  • Normal distribution

  • pnorm(2) shows cdf for standard normal distribution: $$ Z \sim N(0,1) \ \text{,where } \mu = 0 , \sigma^2 = 1\ F(2) = 0.9772 $$

  • pnorm(2,1,1) shows cdf for X:
X \sim N(1,1)\\ \text{,where } \mu = 1 , \sigma^2 = 1\\ F(2) = 0.8413

  • pnorm(2,1,2) shows cdf for Y:

$$ Y \sim N(1,4)\ \text{,where } \mu = 1 , \sigma^2 = 4\ F(2) = 0.6915 $$

Note that the graph has been scaled.

b)

What is the result of the following command: qnorm(pnorm(2))?

pnorm gives "a distance" up to F(2), which is 0.9772499. We can use "that distance" in qnorm function to find a value again, which is 2. (See Figure 2.2 in the book for more)

> pnorm(2)
[1] 0.9772499
> qnorm(pnorm(2))
[1] 2

c)

State what the numbers represent in the three cases.

  • First case

qnorm(0.975)
[1] 1.96
N(0,1) has the 97,5% quantile at x=1.96

  • Second case

qnorm(0.975,1,1)
[1] 2.96
N(1,1) has the 97,5% quantile at x=2.96

  • Third case

qnorm(0.975,1,2)
[1] 4.92
N(1,4) has the 97,5% quantile at x=4.92

2.10 The normal pdf

a)

Which of the following statements regarding the probability density function of the normal distribution N(1, 2^2) is false?

  1. The total area under the curve is equal to 1.0
  2. The mean is equal to 1^2
  3. The variance is equal to 2.
  4. The curve is symmetric about the mean.
  5. The two tails of the curve extend indefinitely.

Answer: 3 is false. Variance is equal to 4, because the standard deviation is equal to 2:

N(1,2^2) \text{,where }N(\mu, \sigma^2) \\ \sigma \text{- standart deviation} \\ \sigma^2 \text{- variance}

b)

Let X be normally distributed with mean 24 and variance 16. Calculate the following probabilities:

P(X\leqslant20) = 0.1586553 
>   pnorm(20, mean=24, sd=sqrt(16) )
[1] 0.1586553
P(X>29.5) = 1 - P(X\leqslant29.5) = 0.08456572 
>  1 - pnorm(29.5, mean=24, sd=sqrt(16) )
[1] 0.08456572
P(X=23.8) = 0.09961098 
> dnorm(23.8, mean=24, sd=sqrt(16))
[1] 0.09961098

2.11 Computer chip control

A machine for checking computer chips uses on average 65 milliseconds per check with a standard deviation of 4 milliseconds. A newer machine, potentially to be bought, uses on average 54 milliseconds per check with a standard deviation of 3 milliseconds. It can be used that check times can be assumed normally distributed and independent.

a)

What is the probability that the time savings per check using the new machine is less than 10 milliseconds is?

By theorem 2.40 and example 2.41:

X \sim N(65,16) \\ Y \sim N(54,9) \\ Z = X - Y \\ \mu_{Z} =\mu_{X} - \mu_{Y} = 65-54=11 \\ \sigma^2_{Z} = \sigma^2_{X}+\sigma^2_{Y}= 16+9=25\\ Z\sim N(11,25)\\ P(Z\leqslant10)=0.4207403 
> pnorm(10, 11,sqrt(25))
[1] 0.4207403

Note that it does not matter in continuous distributions if we use < or \leqslant in order to define the probability.

b)

What is the mean and standard deviation for the total time use for checking 100 chips on the new machine is?

By theorem 2.56:

Y \sim N(54,9) \\ \mu_{Z}= 100*E(X)= 100*54= 5400 \\ \sigma_{Z} = \sqrt{V(100X)} = \sqrt{100*9} = 30

2.12 Concrete items

A manufacturer of concrete items knows that the length (L) of his items are reasonably normally distributed with \mu_{L} = 3000 mm and \sigma_{L} = 3 mm. The requirement for these elements is that the length should be not more than 3007 mm and the length must be at least 2993 mm.

a)

The expected error rate in the manufacturing will be?

  • The expected error rate is the probability of manufacturing an item shorter than 2993 and longer than 3007, so:
L \sim (3000, 9)\\
P(L>3007) = 1 - P(L<3007) \\
P(2993>L>3007) = P(L>3007) + P(L<2993) = \\ (1 - P(L<3007)) + P(L<2993) = \\ (1-0.9901847) + 0.009815329 = 0.01963066 
> pnorm(3007,mean=3000, sd=sqrt(9))
[1] 0.9901847
> pnorm(2993,mean=3000, sd=sqrt(9))
[1] 0.009815329

b)

The concrete items are supported by beams, where the distance between the beams is called L_{beam} and can be assumed normal distributed. The concrete items length is still called L. For the items to be supported correctly, the following requirements for these lengths must be fulfilled: 90mm < L - L_{beam} < 110 mm. It is assumed that the mean of the distance between the beams is \mu_{beam} = 2900 mm. How large may the standard deviation \sigma_{beam} of the distance between the beams be if you want the requirement fulfilled in 99% of the cases?

  • Find:
\sigma_{beam}
  • Needs be fulfilled:
P(90<L-L_{beam}<110) = 0.99
  • We know:
L\sim(3000,9)\\ L_{beam} \sim (2900, \sigma^2_{beam}) \\ \mu_{L-L_{beam}}=\mu_{L}-\mu_{beam}=3000-2900=100\\ \sigma_{L-L_{beam}}=\sqrt{9+\sigma^2_{beam}}\\
  • We can draw a graph like this, because 99% of all distances should between 90 and 110. So there is only 0.5% left in each side, that the 0.005 quantile is at 90 and the 0.995 quantile is at 110:

  • We can also write this like following:
P(90<L-L_{beam}<110) = P(L-L_{beam}<110) - P(L-L_{beam}<90) \\ P(L-L_{beam}<110) = 0.995 \\ P(L-L_{beam}<90) = 0.005 \\
  • We can use theorem 2.43 in order to find a standard deviation of the distance, because the value of the standardized normal random variable at 0.005 quantile should be equal to the transformed distance variable:
z_{0.005} = \frac{90-100}{\sqrt{9+\sigma^2_{beam}}}
  • And the same at 0.995 quantile:
z_{0.995} = \frac{110-100}{\sqrt{9+\sigma^2_{beam}}}
  • We choose the second equation:
z_{0.995}= 2.575829 \\ 2.575829 = \frac{110-100}{\sqrt{9+\sigma^2_{beam}}} 
> qnorm(0.995)
[1] 2.575829
  • So the result is:
\sigma_{beam} = 2.464107

2.13 Online statistic video views

In 2013, there were 110,000 views of the DTU statistics videos that are available online. Assume first that the occurrence of views through 2014 follows a Poisson process with a 2013 average: \lambda_{365days} = 110000.

a)

What is the probability that in a randomly chosen half an hour there is no occurrence of views?

Here we can use either Exponential distribution or Poisson to find the probability.
- Poisson distribution

\lambda_{30min}= \frac{110000*30}{365*24*60} = 6.278539 \\ X \sim Po(\lambda_{30min}= 6.278539 ) \\ P(X=0) = 0.00187614 
> dpois(x=0,  lambda=6.278539 )
[1] 0.00187614
  • Exponential distribution
X \sim Exp(\lambda_{365days}=110000) \\

0 event between now and 30 min is given by: $$ P(X>\frac{30}{365*24*60}) = $$

or can be written as 1 minus the probability of occurring next event between now and 30min $$ = 1 - P(X\leqslant\frac{30}{365*24*60}) = 0.00187614 $$

> 1-pexp(q=30/(365*24*60), rate = 110000)
[1] 0.00187614

b)

There has just been a view, what is the probability that you have to wait more than fifteen minutes for the next view?

  • 0 events between now and 15min is given by:
P(X>\frac{15}{365*24*60}) = \\ 1 - P(X\leqslant\frac{15}{365*24*60}) = 0.04331443 
> 1-pexp(q=15/(365*24*60), rate = 110000)
[1] 0.04331443