Question

Suppose that at a hotel's registration desk there are 5 guests who check-in and need their luggage delivered to their rooms. Suppose that the probability that the luggage will be delivered within five minutes is 0.41, and that you would like to know how likely certain outcomes are, based on those assumptions.

Please think for a moment about what framework you would use to model this....

1. Are thre n identical trials?   (Click to select)  Yes  No
2. Does each trial result in either a "success" or "failure?"   (Click to select)  Yes  No
3. Does the probability of success, p, remain constant?   (Click to select)  Yes  No
4. Is each of the trials independent?   (Click to select)  Yes  No

True or False:

Based on criteria 1~4 above, we can use a member of the family of binomial distributions to model this problem.

(Click to select)  True  False

You should come to the conclusion that you could let T (on Time) be a random variable that counts the number of times that that the luggage is delivered on time. Then T is a binomial random variable with:

• n =  ,
• p =  , and
• q =  .

Using the notation B(n, p) for a binomial random variable with n trials with a probability of "success" of p, we have:

T ~ B(5, p = 0.41)

(b) For each value of s, calculate P(T=s). (Round final answers to 4 decimal places.)

s	P(T = s)
5
4
3
2
1
0

NOTE: You may use the Excel function =BINOM.DIST(number_s, trials, prob_s, cumulative) where

• number_s = # of successes which is t in the table above
• trials = # of trials, n
• prob_s = the probability of success, p
• cumulative = TRUE to return the CDF & FALSE to return the PMF ; set FALSE to return PMF.

NOTE2: Even if you use the Excel function: YOU SHOULD BE SURE THAT YOU CAN COMPUTE TWO OR THREE OF THE VALUES DIRECTLY (i.e. USING THE FORMULA):

P(T=s) = ₅C_s p^s q^5-s

(c) What is the likelihood that exactly three sets of luggage are delivered on time? Find P(T = 3). (Round final answer to 4 decimal places.)

P(T=3) =          

(d) What is the likelihood that up to three sets of luggage are delivered on time? Find P(T ≤ 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(T ≤ 3) =          

(e) How likely are up to two sets of luggage delivered on time? Find P(T < 3). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(T < 3) = P(T ≤ 2) =              

(f) How likely is it that four or more sets of luggage are delivered on time? Find P(T ≥ 4). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(x ≥ 4) =          

(g) How likely is it that more than two sets of luggage will be delivered on time? Find P(T > 2). (Do not round intermediate calculations. Round final answer to 4 decimal places.)

P(T > 2) =             

(h) Use the probabilities you computed in part b to calculate the mean μ_T, the variance, σ ²_T , and the standard deviation, σ_T, of this binomial distribution. Show that the formulas for μ_T ,σ ²_T, and σ_T given in this section give the same results. (Do not round intermediate calculations. Round final answers to µ_T in to 2 decimal places, σ ²_T and σ_T in to 4 decimal places.)

µT
σ2Tσ2T
σT

(i) Calculate the interval [μ_T ± 2σ_T]. Use the probabilities of part b to find the probability that t will be in this interval. Hint: When calculating probability, round up the lower interval to next whole number and round down the upper interval to previous whole number. (Round your answers to 4 decimal places. A negative sign should be used instead of parentheses.)

The interval is [  ,  ].

P(  ≤ t ≤  ) =

Answer 1

Based on the given data,

1. Are thre n identical trials? Yes ( Event that guest checks in - happens n = 5 times)
2. Does each trial result in either a "success" or "failure?" Yes (Succes being the event that that the luggage will be delivered to the guest within five minutes)
3. Does the probability of success, p, remain constant?   Yes (p=0.41 in each trial)
4. Is each of the trials independent?   Yes  

True or False:

Based on criteria 1~4 above, we can use a member of the family of binomial distributions to model this problem.

True  

T is a binomial random variable with:

• n =  ,5
• p = 0.41, and
• q = 1 - 0.41 = 0.59

Using the notation B(n, p) for a binomial random variable with n trials with a probability of "success" of p, we have:

(b) T ~ B(5, p = 0.41)

For a Binomial random variable, the probability mass function:

Using excel function "BINOM.DIST"

We get:

S	Probability
5	0.0116
4	0.0834
3	0.2399
2	0.3452
1	0.2484
0	0.0715

(c)The likelihood that exactly three sets of luggage are delivered on time

(d) The likelihood that up to three sets of luggage are delivered on time

= 0.0715 + 0.2484 + 0.3452 + 0.2399

= 0.9051

(e) The likelihood that up to three sets of luggage are delivered on time

  = 0.0715 + 0.2484 + 0.3452

= 0.6651

(f) Probability that four or more sets of luggage are delivered on time

= 0.0834 + 0.0116

= 0.0949

(g) Probability that more than two sets of luggage will be delivered on time

= 1 - 0.6651

= 0.3349

(f) Finding the expectation:

S	Probability	SxPr
5	0.0116	0.058
4	0.0834	0.333
3	0.2399	0.720
2	0.3452	0.690
1	0.2484	0.248
0	0.0715	0.000
	Expectation	2.05

By formula,

E(T) np = (5) (0.41) = 2.05

S	Probability	S2 x Pr
5	0.0116	0.290
4	0.0834	1.334
3	0.2399	2.159
2	0.3452	1.381
1	0.2484	0.248
0	0.0715	0.000
	SUM	5.412

Variance = S2 P(T) - E(T)2

= 5.412 - (2.05)2

= 1.2095

By formula:

Variance = n p q = (5)(0.41)(0.59) = 1.2095

Standard deviation

By formula:

Standard deviation

(i)  [μT ± 2σT]

= [2.05 ± 2(1.10)]

= -0.1495, 4.2495

The interval is [*0.1495, 4.2495].

The probability that T lies within the interval (-0.1495, 4.2495)

= P(T=0) + P(T=1) + P(T=2) + P(T = 3) + P(T = 4)

=0.0715+0.2484+0.3452+0.2399+0.0834

= 0.9884