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Using Continuous Probability Distribution Determine X 7

1. Let x be the random variable described by the uniform probability distribution with its lower bound at a = 120, upper bound at b = 140.

(a) What is the probability density function, f(x)?

Answer:if a = 120 and b = 140, then

fig

(b) What is E(x) and σ?

Answer:the expected value and standard deviation arefig

2. What is p(x = 130)? Explain why p(x = 130) ≠ 1/20.

Answer: p(x = 130) = 0, just as it is for any value of x in the interval 120 ≤ x ≤ 140. The probability density function f(x) does not provide probabilities but only the height of the curve above the horizontal axis at a particular value of x. In this case, the height (but not the probability) is 1/20.

3. Referring to the previous exercise, find the following probabilities using f(x) and R.

(a) p(125≤ x ≤ 135)

Answer: 0.5000

fig

#Comment. Subtract punif(125,min=120,max=140) from
#punif(135,min=120,max=140).
punif(135, min = 120, max = 140) -
punif(125, min = 120, max = 140)

## [1] 0.5

(b) p(125 ≤ x ≤  131)

Answer:0.3000

fig

#Comment. Subtract punif(125,min=120,max=140) from
#punif(131,min=120,max=140).
punif(131, min = 120, max = 140) -
punif(125, min = 120, max = 140)
##   [1]   0.3

(c) p(129 ≤ x ≤  131)

Answer:0.1000

fig

#Comment. Subtract punif(129,min=120,max=140) from
#punif(131,min=120,max=140).
punif(131, min = 120, max = 140) -
punif(129, min = 120, max = 140)
##   [1]   0.1

(d) p(120.50 ≤ x ≤139.50)

Answer:0.9500

fig

#Comment. Subtract punif(120.50,min=120,max=140) from
#punif(139.50,min=120,max=140).
punif(139.50, min = 120, max = 140) -
punif(120.50, min = 120, max = 140)
##   [1]   0.95

4. Referring to the previous exercise, find the following probabilities using f(x) and R.

(a) p(x ≥ 124)

Answer: 0.8000

fig

#Comment1. 1 minus punif(124,min=120,max=140)
1 - punif(124, min = 120, max = 140)
##   [1]   0.8
#Comment2. Use punif(124,min=120,max=140,lower.tail=FALSE);
#the argument lower.tail=FALSE results in the upper tail area,
#not the lower tail area (as normally happens with punif()).

punif(124, min = 120, max = 140, lower.tail = FALSE)
##  [1]   0.8

(b) p(x ≥ 128)

Answer: 0.6000

fig

#Comment1. 1 minus punif(128,min=120,max=140).
1 - punif(128, min = 120, max = 140)
##   [1]   0.6
#Comment2. Use punif(128,min=120,max=140,lower.tail=FALSE).
punif(128, min = 120, max = 140, lower.tail = FALSE)
##   [1]   0.6

(c) p(x ≥ 134)

Answer: 0.3000

fig

#Comment1. 1 minus punif(134,min=120,max=140).
1 - punif(134, min = 120, max = 140)
## [1] 0.3
#Comment2. Use punif(134,min=120,max=140,lower.tail=FALSE).
punif(134, min = 120, max = 140, lower.tail = FALSE)
## [1] 0.3

(d) p(x ≥ 139)

Answer:0.0500

fig

#Comment1. 1 minus punif(139,min=120,max=140).
1 - punif(139, min = 120, max = 140)
##   [1]   0.05
#Comment2. Use punif(139,min=120,max=140,lower.tail=FALSE).
punif(139, min = 120, max = 140, lower.tail = FALSE)
##   [1]   0.05

5. A recent college graduate is moving to Houston, Texas to take a new  Job, and is looking to purchase a home. Since Greater Houston takes in a relatively large metropolitan area of nearly 7,000,000  people, there are many homes from which to choose. When consulting the real estate web sites, it is possible to select the price range of housing in which one is most interested. Suppose the potential buyer species $200,000 to $250,000, and the result returns 105 homes with prices distributed uniformly throughout that range. Please answer the following questions

(a) What would be the probability density function that best describes this distribution of housing prices?

Answer:

fig

(b) What are E(x) and σ?

Answer:

fig

(c) If the buyer ultimately selects a home randomly from this initial list of 105 homes, what is the probability she will have to pay more than $235,000?

Answer: 0.3000

fig

#Comment1. 1 minus punif(235000,min=200000,max=250000).
1 - punif(235000, min = 200000, max = 250000)
##  [1]   0.3
#Comment2. Or punif( ,lower.tail=FALSE)
punif(235000, min = 200000, max = 250000, lower.tail = FALSE)
##  [1]   0.3

6. Referring to the above question, what is the minimum price the buyer would pay if she refines her focus to the upper 25% of the price range from $200,000 to $250,000? Note that it is possible to omit the min= and max= notation as long as the arguments are placed in the same order; that is, as long as the lower bound is placed first, the upper bound is placed immediately after it.

Answer:$237,500

#Comment1. Use function qunif(0.75,200000,250000) to find that price
#for which 25 percent of listings are higher, 75 percent are lower.
qunif(0.75, 200000, 250000)
##  [1]   237500
#Comment2. Use qunif(0.25, 200000, 250000, lower.tail = FALSE)
qunif(0.25, 200000, 250000, lower.tail = FALSE)
##  [1]   237500

7. Use R to answer the following questions concerning z, the standard normal variable.

(a) p(z ≤ 2.33)?

Answer: 0.9901. The probability that z is less than or equal to 2.33 is about 0.99.

#Comment. Use function pnorm(2.33) to find the probability
#that z will be less than or equal to 2.33.
pnorm(2.33)
##   [1]   0.9900969

(b) p(z ≤ 2.05)?

Answer: 0.9798. The probability that z is less than or equal to 2.05 is about 0.98.

#Comment. Use function pnorm(2.05) to find the probability
#that z will be less than or equal to 2.05.
pnorm(2.05)
##   [1]   0.9798178

 (c) p(z ≤ 1.96)?

Answer: 0.9750. The probability that z is less than or equal to 1.96 is about 0.9750.

#Comment. Use function pnorm(1.96) to find the probability
#that z will be less than or equal to 1.96.
pnorm(1.96)
##   [1]  0.9750021

(d) p(z ≤ 1.28)?

Answer: 0.8997. The probability that z is less than or equal to 1.28 is about
0.90.

#Comment. Use function pnorm(1.28) to find the probability
#that z will be less than or equal to 1.28.

pnorm(1.28)
##  [1]   0.8997274

8. Use R to answer the following questions about z.

(a) p(z ≤ 0.00)?

Answer: 0.5000. The probability that z will be less than or equal to 0.00 is 0.5000.

#Comment. Use function pnorm() to find cumulative probability.
pnorm(0.00)
##   [1]   0.5

(b) p(z ≤ –1.28)?

Answer: 0.1003. The probability that z will be less than or equal to -1.28 is about 0.10.

#Comment. Use function pnorm() to find cumulative probability.
pnorm(-1.28)
##   [1]   0.1002726

(c) p(z ≤ –1.96)?

Answer: 0.025. The probability that z will be less than or equal to -1.96 is about 0.025.

#Use function pnorm() to find cumulative probability.
pnorm(-1.96)
##   [1]   0.0249979

(d) p(z ≤ –2.05)?

Answer: 0.02018. The probability that z will be less than or equal to -2.05 is about 0.02.

#Comment. Use function pnorm() to find cumulative probability.
pnorm(-2.05)
##   [1]   0.02018222

(e) p(z ≤ –2.33)?

Answer: 0.009903. The probability that z will be less than or equal to -2.33 is about 0.01.

#Comment. Use function pnorm() to find cumulative probability.
pnorm(-2.33)
##   [1]   0.009903076

 9. Use R to answer the following questions about z.

(a) p(z ≥ −2.33)

Answer: 0.9901. The probability that z will be greater than or equal to -2.33 is about 0.99.

#Comment1. 1 minus probability z is less than or equal to -2.33.

1 pnorm(2.33)

## [1] 0.9900969

#Comment2. Use pnorm(−2.33, lower.tail = FALSE) to confirm.

pnorm(2.33, lower.tail = FALSE)

## [1] 0.9900969

(b) p(z≥ −2.05)

Answer: 0.9798. The probability that z will be greater than or equal to 2.05 is about 0.98.

#Comment1. 1 minus probability z is less than or equal to −2.05.

1 pnorm(2.05)

## [1] 0.9798178

#Comment2. Use pnorm(−2.05, lower.tail = FALSE) to confirm.

pnorm(2.05, lower.tail = FALSE)

## [1] 0.9798178

(c) p(z ≥ −1.96)

Answer: 0.975. The probability that z will be greater than or equal to 1.96 is about 0.975.

#Comment1. 1 minus probability z is less than or equal to −1.96.

1 - pnorm(-1.96)

##  [1] 0.9750021

#Comment2. Use pnorm(-1.96, lower.tail = FALSE) to confirm.

pnorm(-1.96,  lower.tail = FALSE)

## [1] 0.9750021

(d) p(z ≥ −1.28)

Answer: 0.8997. The probability that z will be greater than or equal to -1.28 is about 0.90.

#Comment1. 1 minus probability z is less than or equal to -1.28.

1 - pnorm(-1.28)

##  [1] 0.8997274

#Comment2. Use pnorm(-1.28, lower.tail = FALSE) to confirm.

pnorm(-1.28, lower.tail = FALSE)

## [1] 0.8997274

10. Use R to answer the following questions about z.

(a) p(z ≥ 1.28)

Answer: 0.1003. The probability that z will be greater than or equal to 1.28 is about 0.10.

#Comment1. 1 minus probability z is less than or equal to 1.28.

1 - pnorm(1.28)

## [1] 0.1002726

#Comment2. Use pnorm(1.28, lower.tail = FALSE) to confirm.

pnorm(1.28,  lower.tail  = FALSE)

## [1] 0.1002726

(b) p(z ≥ 1.96)

Answer: 0.025. The probability that z will be greater than or equal to 1.96 is about 0.025.

#Comment1. 1 minus probability z is less than or equal to 1.96.

1 - pnorm(1.96)

## [1] 0.0249979

#Comment2. Use pnorm(1.96, lower.tail = FALSE) to confirm.

pnorm(1.96,  lower.tail  = FALSE)

## [1] 0.0249979

(c) p(z ≥ 2.05)

Answer: 0.02018. The probability that z will be greater than or equal to 2.05 is about 0.02.

#Comment. 1 minus probability z is less than or equal to 2.05.

1 - pnorm(2.05)

## [1] 0.02018222

#Comment2. Use pnorm(2.05, lower.tail = FALSE) to confirm.

pnorm(2.05,  lower.tail  = FALSE)

## [1] 0.02018222

(d) p(z ≥ 2.33)

Answer: 0.009903. The probability that z will be greater than or equal to 2.33 is about 0.01.

#Comment1. 1 minus probability z is less than or equal to 2.33.

1 - pnorm(2.33)

## [1] 0.009903076

#Comment2. Use pnorm(2.33, lower.tail = FALSE) to confirm.

pnorm(2.33, lower.tail  = FALSE)

## [1] 0.009903076

11. Use R to answer the following questions about z.

(a) If the area to the left of z is 0.99, what is z?

Answer: z = 2.326

#Comment. Use function qnorm(0.99) to find value of z

#providing an area of 0.99 to its left.

qnorm(0.99)

## [1] 2.326348

(b) If the area to the left of z is 0.975, what is z?

Answer: z = 1.96

#Comment. Use function qnorm(0.975) to find value of z

#providing an area of 0.975 to its left.

qnorm(0.975)

## [1] 1.959964

(c) If the area to the left of z is 0.95, what is z?

Answer: z = 1.645

#Comment. Use function qnorm(0.95) to find value of z

#providing an area of 0.95 to its left.

qnorm(0.95)

## [1] 1.644854

(d) If the area to the left of z is 0.90, what is z?

Answer: z = 1.282

#Comment. Use function qnorm(0.90) to find value of z

#providing an area of 0.90 to its left.

qnorm(0.90)

## [1] 1.281552

12. Use R to answer the following questions about z.

(a) If the area to the right of z is 0.10, what is z?

Answer: z = 1.282

#Comment1. Use function qnorm(0.90) to find value of z providing

# an area of 0.90 to the left, or what is the same thing an area

# to the right of 0.10.

qnorm(0.90)

## [1] 1.281552

#Comment2. Use qnorm(0.10,lower.tail=FALSE) to confirm.

qnorm(0.10,  lower.tail  =  FALSE)

## [1] 1.281552

(b) If the area to the right of z is 0.05, what is z?

Answer: z = 1.645

#Comment1. Use function qnorm(0.95) to find value of z providing

#an area of 0.95 to the left, or what is the same thing an area

#to the right of 0.05.

qnorm(0.95)

## [1] 1.644854

#Comment2. Use qnorm(0.05, lower.tail = FALSE) to confirm.

qnorm(0.05,  lower.tail  =  FALSE)

## [1] 1.644854

(c) If the area to the right of z is 0.025, what is z?

Answer: z = 1.96

#Comment1. Use function qnorm(0.975) to find z providing

# an area of 0.975 to the left, or what is the same thing an area

# to the right of 0.025.

qnorm(0.975)

## [1] 1.959964

#Comment2. Use qnorm(0.025, lower.tail = FALSE) to confirm.

qnorm(0.025,  lower.tail  =  FALSE)

 ## [1] 1.959964

(d) If the area to the right of z is 0.01, what is z?

Answer: z = 2.326

#Comment1. Use function qnorm(0.99) to find value of z providing

# an area of 0.99 to the left, or what is the same thing an area

#to the right of 0.01.

qnorm(0.99)

## [1] 2.326348

#Comment2. Use qnorm(0.01, lower.tail = FALSE) to confirm.

qnorm(0.01,  lower.tail  =  FALSE)

## [1] 2.326348

13. Let x be a normally-distributed random variable with a mean µ = 25 and standard deviation σ = 5, answer the following questions.

(a) What is the probability that x will be less than or equal to 35?

Answer: 0.9772

13_eq1

#Comment. Use pnorm(35,25,5) for probability x is less than

#or equal to 35 when distribution has mean of 25 and standard

# devation of 5.

pnorm(35,  25, 5)

##  [1] 0.9772499

(b) What is the probability that x will be less than or equal to 32?

Answer: 0.9192

13b_eq1

#Comment. Use pnorm(32,25,5) for probability x is less than

#or equal to 32 when distribution has mean of 25 and standard

#devation of 5.

pnorm(32, 25, 5)

## [1] 0.9192433

(c) What is the probability that x will be less than or equal to 30?

Answer: 0.8413

13c_eq1

#Comment. Use pnorm(30,25,5) for probability is less than

#or equal to 30 when distribution has mean of 25 and standard

#devation of 5.

pnorm(30, 25, 5)

##  [1] 0.8413447

(d) What is the probability that x will be less than or equal to 25?

Answer: 0.50

13d_eq1

#Comment. Use pnorm(25,25,5) for probability x is less than

#or equal to 25 when distribution has mean of 25 and standard

#devation of 5.

pnorm(25, 25, 5)

## [1] 0.5

14. If x is a normally-distributed random variable with a mean µ = 63 and standard deviation σ = 4.5, answer the following questions.

(a) What is the probability that x will be less than or equal to -57?

Answer: 0.9088

14a_eq.png

#Comment. Use pnorm(-57,-63,4.5) for probability x is less than

#or equal  to  -57  when  distribution  has  mean  of  -63  and standard

#devation of 4.5.

pnorm(-57, -63, 4.5)

## [1] 0.9087888

(b) What is the probability that x will be less than or equal to -60?

Answer: 0.7475

14b_eq1

#Comment. Use pnorm(-60,-63,4.5) for probability x is less than

#or  equal  to  -60  when  distribution  has  mean  of  -63  and standard

#devation of 4.5.

pnorm(-60, -63, 4.5)

## [1] 0.7475075

(c) What is the probability that x will be less than or equal to -63?

Answer: 0.50

14c_eq1

#Comment. Use pnorm(-63,-63,4.5) for probability x is less than

#or  equal  to  -63  when  distribution  has  mean  of  -63  and standard

#devation of 4.5.

pnorm(-63, -63, 4.5)

## [1] 0.5

(d) What is the probability that x will be less than or equal to -70?

Answer: 0.05991

14d_eq1

#Comment. Use pnorm(-70,-63,4.5) for probability x is less than

#or  equal  to  -70  when  distribution  has  mean  of  -63  and standard

#devation of 4.5.

pnorm(-70, -63, 4.5)

## [1] 0.05990691

15. If x is a normally-distributed random variable with a mean µ = 36 and standard deviation σ = 3, answer the following questions.

(a) What is the probability that x will be greater than 42?

Answer: 0.02275

15a_eq1

#Comment1. 1 minus probability x is less than or equal to 42

#when distribution has mean of 36 and standard deviation of 3.

1  -  pnorm(42, 36, 3)

## [1] 0.02275013

#Comment2. Use pnorm(42,36,3,lower.tail=FALSE) to confirm.

pnorm(42, 36, 3,  lower.tail  =  FALSE)

## [1] 0.02275013

(b) What is the probability that x will be greater than 39?

Answer: 0.1587

15b_eq1

#Comment1. 1 minus probability x is less than or equal to 39

#when distribution has mean of 36 and standard deviation of 3.

1  -  pnorm(39, 36, 3)

## [1] 0.1586553

#Comment2. Use pnorm(42,36,3,lower.tail=FALSE) to confirm.

pnorm(39, 36, 3, lower.tail  =  FALSE)

## [1] 0.1586553

(c) What is the probability that x will be greater than 33?

Answer: 0.8413

15c_eq1

#Comment1. 1 minus probability x is less than or equal to 33

#when distribution has mean of 36 and standard deviation of 3.

1  -  pnorm(33, 36, 3)

##  [1]  0.8413447

#Comment2. Use pnorm(33,36,3,lower.tail=FALSE) to confirm.

pnorm(33, 36, 3, lower.tail = FALSE)

##  [1]  0.8413447

(d) What is the probability that x will be greater than 27?

Answer: 0.9987

15d_eq1

#Comment1. 1 minus probability x is less than or equal to 27

#when distribution has mean of 36 and standard deviation of 3.

1 - pnorm(27, 36, 3)

## [1] 0.9986501

#Comment2. Use pnorm(27,36,3,lower.tail=FALSE) to confirm.

pnorm(27,  36,  3,  lower.tail  =  FALSE)

## [1] 0.9986501

16. Let x be a normally-distributed random variable with a mean of µ = 100 and a standard deviation of σ = 15.

(a) If the area to the left of x is 0.99, what is x?

Answer: x = 134.9

Method 1. Recall

16a_eq1

and the area to the left of z is 0.99 when

z = 2.326

Substituting and solving for x

16a_eq3

Plugging in values for the mean and standard deviation µ = 100 and σ = 15

x = µ + 2.326σ = 100 + (2.326)(15) = 100 + 34.89 = 134.9

Method 2. Use R.

#Comment1. Use qnorm(0.99,100,15) to find value of x providing

#area  of  0.99  to  left  with  mean  100  and  standard  deviation 15.

qnorm(.99, 100, 15)

## [1] 134.8952

#Comment2. Use qnorm(0.01,100,15,lower.tail=FALSE) to confirm.

qnorm(0.01, 100, 15, lower.tail = FALSE)

## [1] 134.8952

(b) If the area to the left of x is 0.975, what is x?

Answer: x = 129.4

#Comment1. Use qnorm(0.975,100,15) to find value of x providing

#area  of  0.975  to  left  with  mean  100  and  standard  deviation 15.

qnorm(.975, 100, 15)

## [1] 129.3995

#Comment2. Use qnorm(0.025,100,15,lower.tail=FALSE) to confirm.

qnorm(0.025, 100, 15, lower.tail = FALSE)

## [1] 129.3995

(c) If the area to the left of x is 0.95, what is x?

Answer: x = 124.7

#Comment1. Use qnorm(0.95,100,15) to find value of x providing

#area  of  0.95  to  left  with  mean  100  and  standard  deviation 15.

qnorm(.95, 100, 15)

## [1] 124.6728

#Comment2. Use qnorm(0.05,100,15,lower.tail=FALSE) to confirm.

qnorm(0.05, 100, 15, lower.tail = FALSE)

## [1] 124.6728

(d) If the area to the left of x is 0.90, what is x?

Answer: x = 119.20

#Comment1. Use qnorm(0.90,100,15) to find value of x providing

# area  of  0.90  to  left  with  mean  100  and  standard  deviation 15.

qnorm(.90, 100, 15)

## [1] 119.2233

#Comment2. Use qnorm(0.10,100,15,lower.tail=FALSE) to confirm.

qnorm(0.10, 100, 15, lower.tail = FALSE)

## [1] 119.2233

17. Let xbe a normally-distributed random variable with a mean of µ = 280 and a standard deviation of σ = 35.

(a) If the area to the right of x is 0.10, what is x?

Answer: x = 235.1

Method 1. Recall

17a_eq1

and the area to the right of x is 0.10 when

z = 1.28

Substituting and solving for x

17a_eq2

Plugging in the values for the mean and standard deviation µ = −280 andσ = 35

x = µ + 1.28σ = 280 + (1.28)(35) = 280 + 44.8 = 235.1

Method 2. Use R.

#Comment1. Use qnorm(0.90,-280,35) to find value of x providing

#area  of  0.90  to  left  with  mean  -280  and  standard  deviation  35.

qnorm(0.90, -280, 35)

## [1] -235.1457

#Comment2. Use qnorm(0.10,-280,35,lower.tail=FALSE) to confirm.

qnorm(0.10, -280, 35, lower.tail = FALSE)

## [1] -235.1457

(b) If the area to the right of x is 0.05, what is x?

Answer: x = −222.4

#Comment1. Use qnorm(0.95,-280,35) to find value of x providing

#area  of  0.95  to  left  with  mean  -280  and  standard  deviation  35.

qnorm(0.95, -280, 35)

## [1] −222.4301

#Comment2. Use qnorm(0.05,-280,35,lower.tail=FALSE) to confirm.

qnorm(0.05, −280, 35, lower.tail = FALSE)

## [1] −222.4301

(c) If the area to the right of x is 0.025, what is x?

Answer: x = 211.4

#Comment1. Use qnorm(0.975,-280,35) to find value of x providing

#area  of  0.975  to  left  with  mean  -280  and  standard  deviation  35.

qnorm(0.975, -280, 35)

## [1] -211.4013

#Comment2. Use qnorm(0.025,-280,35,lower.tail=FALSE) to confirm.

qnorm(0.025, -280, 35, lower.tail = FALSE)

## [1] -211.4013

(d) If the area to the right of x is 0.01, what is x?

Answer: x = 198.6

#Comment1. Use qnorm(0.99,-280,35) to find value of x providing

#area  of  0.99  to  left  with  mean  -280  and  standard  deviation  35.

qnorm(0.99, -280, 35)

## [1] -198.5778

#Comment2. Use qnorm(0.01,-280,35,lower.tail=FALSE) to confirm.

qnorm(0.01, -280, 35, lower.tail = FALSE)

## [1] -198.5778

18. According to British weather forecasters, the average monthly rainfall in London during the month of June is µ = 2.09 inches. Assume the monthly precipitation is a normally-distributed random variable with a standard deviation of σ = 0.48 inches.

(a) What is the probability that London will have between 1.5 and 2.5 inches of precipitation next June?

Answer: 0.694

figure

#Comment. pnorm(2.5,2.09,0.48) less pnorm(1.5,2.09,0.48)

pnorm(2.5, 2.09, 0.48) - pnorm(1.5, 2.09, 0.48)

## [1] 0.693989

(b) What is the probability that London will have 1 inch or less of precipitation?

Answer: 0.01158

18b_eq1

#Comment. Use pnorm(1,2.09,0.48) to find probability.

pnorm(1, 2.09, 0.48)

## [1] 0.01157853

(c) If London authorities prepare for flood conditions when the monthly precipitation falls in the upper 5% of the normal June amounts, how much rain would have to fall to cause local authorities to begin flood preparations?

Answer: x = 2.88

Since z = 1.645 cuts off the upper 5% of the standard normal probability distribution, we solve for the corresponding value of x

18c_eq1

x = µ + (1.645)σ = 2.09 + (1.645)(0.48) = 2.09 + 0.79 = 2.88

#Comment1. Use qnorm(0.95,2.09,0.48) to find x providing area of 0.95 to left with mean 2.09 and standard deviation 0.48

qnorm(.95, 2.09, 0.48)

## [1] 2.87953

#Comment2. Use qnorm(0.05,2.09,0.48,lower.tail=FALSE) to confirm

qnorm(0.05, 2.09, 0.48, lower.tail = FALSE)

## [1] 2.87953

19. The time required by students enrolled in a pre-medical program to complete an organic chemistry final examination is normally-distributed with a mean of µ = 200 minutes and standard deviation of σ = 20 minutes.

(a) What is the probability a student will complete the examination in 180 minutes or less?

Answer: 0.1587

19a_eq1

#Comment. Use pnorm(180,200,20) to find probability.

pnorm(180, 200, 20)

## [1] 0.1586553

(b) What is the probability a student will take between 180 and 220 minutes to complete the examination?

Answer: 0.6827

19b_eq1

#Comment. Subtract pnorm(180,200,20) from pnorm(220,200,20).

pnorm(220, 200, 20) - pnorm(180, 200, 20)

## [1] 0.6826895

(c) Since this particular class is a large lecture section of 300 students, and the final examination period lasts 240 minutes, how many students would we expect to submit the completed exam on time?

Answer: Around 293 to 294 students should finish on time.

19c_eq1

pnorm(240, 200, 20) * 300

## [1] 293.175

20. A large commerical agricultural concern in Spain produces melons with a diameter that is normally-distributed with a mean of µ = 15 centimeters (cm) and a standard deviation of σ = 2 cm.

(a) What is the probability that a melon will have a diameter of at least 12 cm?

Answer: 0.9332

20a_eq1

#Comment1. Use 1 minus pnorm(12,15,2) to find probability.

1 - pnorm(12, 15, 2)

## [1] 0.9331928

#Comment2. Use pnorm(12,15,2,lower.tail=FALSE) to confirm.

pnorm(12, 15, 2, lower.tail = FALSE)

## [1] 0.9331928

(b) What is the probability that a randomly selected melon will have a diameter of no less than 12 cm but no more than 16 cm?

Answer: 0.6247

20b_eq1

#Comment. Subtract pnorm(12,15,2) from pnorm(16,15,2).

pnorm(16, 15, 2) - pnorm(12, 15, 2)

## [1] 0.6246553

(c) The producer has an arrangement with one retailer by which they receive a slightly higher price for melons with a diameter falling in the top 10%. What is the minimum diameter a melon must have in order to qualify for the higher price?

Answer: x = 17.56 cm

Since z = 1.28 cuts off the upper 10% of the standard normal probability distribution, we solve for the corresponding value of x

20c_eq1

x = µ + (1.28)σ = 15 + (1.28)(2) = 15 + 2.56 = 17.56

Use R to confirm that a melon with a diameter of 17.56 cm sets the cut-off point for the top 10%.

#Comment1. Use qnorm(.90,15,2) to find value of x providing area of 0.90 to left with mean 15 and standard deviation 2.

qnorm(.90, 15, 2)

## [1] 17.5631

#Comment2. Use qnorm(0.10,15,2,lower.tail=FALSE) to confirm.

qnorm(0.10, 15, 2, lower.tail = FALSE)

## [1] 17.5631

 21. A variable is exponentially-distributed with a mean of µ = 5.

(a) What is the form of the probability density function?

Answer:

21a_eq1

(b) What is the cumulative probability function?

Answer:

21b_eq1

(c) What is p(x ≤ 2)?

Answer: 0.3297

21c_eq1

#Comment. Use pexp(2,1/5) to find probability x is less than or equal to 2 when the mean is 5.

pexp(2,1/5)

## [1] 0.32968

(d) What is p(x ≤ 4)?

Answer: 0.5507

21d_eq1

#Comment. Use pexp(4,1/5) to find probability x is less than

# or equal to 4 when the mean is 5.

pexp(4, 1/5)

## [1] 0.550671

(e) What is p(x ≤ 5)?

Answer: 0.6321

21e_eq1

#Comment. Use pexp(5,1/5) to find probability x is less than

#or equal to 5 when the mean is 5.

pexp(5,1/5)

## [1] 0.6321206

(f) What is p(x ≤ 8)?

Answer: 0.7981

21f_eq1

#Comment. Use pexp(8,1/5) to find probability x is less than

#or equal to 8 when the mean is 5.

pexp(8,1/5)

## [1] 0.7981035

(g) What is p(x ≤ 15)?

Answer: 0.9502

21g_eq1

#Comment. Use pexp(15,1/5) to find probability x is less than

#or equal to 15 when the mean is 5.

pexp(15, 1/5)

## [1] 0.9502129

22. Referring to the preceding exercise, answer the following questions.

(a) What is p(x ≥ 2)?

Answer: 0.6703

22a_eq1

#Comment1. 1 minus probability x is less than or equal to 2 when the mean is 5.

1 - pexp(2, 1/5)

## [1] 0.67032

#Comment2. Use pexp(2,1/5,lower.tail=FALSE) to confirm.

pexp(2, 1/5, lower.tail = FALSE)

## [1] 0.67032

(b) What is p(x ≥ 4)?

Answer: 0.4493

22b_eq1

#Comment1. 1 minus probability x is less than or equal to 4 when the mean is 5.

1 - pexp(4,1/5)

## [1] 0.449329

#Comment2. Use pexp(4,1/5,lower.tail = FALSE) to confirm.

pexp(4, 1/5, lower.tail = FALSE)

## [1] 0.449329

(c) What is p(x ≥ 5)?

Answer: 0.3679

22c_eq1

#Comment1. 1 minus probability x is less than or equal to 5 when the mean is 5.

1 - pexp(5, 1/5)

##  [1] 0.3678794

#Comment2. Use pexp(5,1/5,lower.tail=FALSE) to confirm.

pexp(5, 1/5, lower.tail = FALSE)

## [1] 0.3678794

(d) What is p(x ≥ 8)?

Answer: 0.2019

22d_eq1

#Comment1. 1 minus probability x is less than or equal to 8

#when the mean is 5.

1 - pexp(8, 1/5)

##  [1] 0.2018965

#Comment2. Use pexp(8,1/5,lower.tail=FALSE) to confirm.

pexp(8, 1/5, lower.tail = FALSE)

## [1] 0.2018965

(e) What is p(x ≥ 15)?

Answer: 0.04979

22e_eq1

#Comment1. 1 minus probability x is less than or equal to 15

#when the mean is 5.

1 - pexp(15, 1/5)

##  [1] 0.04978707

#Comment2. Use pexp(15,1/5,lower.tail=FALSE) to confirm.

pexp(15, 1/5, lower.tail = FALSE)

## [1] 0.04978707

23. Referring to the previous exercise, answer the following questions.

(a) What is p(2 ≤ x ≤ 4)?

Answer: 0.221

23a_eq.png

#Comment. Subtract pexp(2,1/5) from pexp(4,1/5).

pexp(4, 1/5) - pexp(2, 1/5)

## [1] 0.2209911

(b) What is p(2 ≤ x ≤ 5)?

Answer: 0.3024

23b_eq1

#Comment. Subtract pexp(2,1/5) from pexp(5,1/5).

pexp(5, 1/5) - pexp(2, 1/5)

## [1] 0.3024406

(c) What is p(2 ≤ x ≤ 8)?

Answer: 0.4684

23c_eq1.png

#Comment. Subtract pexp(2,1/5) from pexp(8,1/5).

pexp(8, 1/5) - pexp(2, 1/5)

## [1] 0.4684235

(d) What is p(2 ≤ x ≤ 15)?

Answer: 0.6205

23d_eq1

#Comment. Subtract pexp(2,1/5) from pexp(15,1/5).

pexp(15, 1/5) - pexp(2, 1/5)

## [1] 0.620533

24. The number of visits to the Book4Less.com discount travel website is a Poisson- distributed random variable with a mean arrival rate of 10 visits per minute.

(a) If the Poisson arrival rate is 10 visitors per minute, what is the mean of the associated exponential probility density function?

Answer: If the mean Poisson arrival rate is 10 visitors per minute (every 60 seconds), then the time between passenger visitors is exponentially-distributed with a mean of 6 seconds.

(b) What is the exponential probability density function?

Answer:

24b_eq1

(c) What is the cumulative probability function?

24c_eq1.png

(d) What is the standard deviation of the distribution?

25. Referring to the preceding exercise, answer the following questions.

(a) If the internet server experiences a brief power failure of 18 seconds duration during which time people would be denied access to the website, what is the probability that no one attempted to visit the Book4Less.com website anyway and thus no business was lost during the down-time? Use the exponential framework to answer this question.

Answer: 0.04979

25a_eq1

#Comment. 1 minus pexp(18,1/6).

1 - pexp(18, 1/6)

## [1] 0.04978707

(b) Answer question (a) using the Poisson probability approach. Confirm that the answer is equal to that in (a).

Answer: 0.04979

Since we are told that there is an average of 10 visitors per minute, or 10 visitors per 60 seconds, we need to convert this parameter: µ=10 visitors/60 seconds=3 visitors/18 seconds.

Recall that the Poisson probability function is

25b_eq1

#Comment.  Use  dpois(0,3)  to  find  probability  of  0 visitors during  a  period  (18  seconds)  when  the  mean is 3 visitors.

dpois(0, 3)

## [1] 0.04978707

(c) Comment on the fact that the answers to parts (a) and (b) are exactly the same regardless of the approach (Poisson or exponential) employed.

Answer: If the interval of time (or distance) between occurrences is distributed exponentially, then the number of occurrences in that interval must be Poisson- distributed. The two distributions are inter-related.

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Source: https://study.sagepub.com/stinerock/student-resources/exercises/chapter-6-continuous-probability-distributions