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mario62 [17]
2 years ago
11

Similar statement of √111 Pls help

Mathematics
1 answer:
Diano4ka-milaya [45]2 years ago
7 0
111 is between 11 and 12
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Support requests arrive at a software company at the rate of 1 every 30 minutes. Assume that the requests arrive as events in a
rewona [7]

Answer:

a. 0.5413

b. 20

c. 0.3724

d. 4.4721

Step-by-step explanation:

Solution:-

- We will start by defining a random variable X.

           

                     X : The number of support requests arrived

- The event defined by the random variable ( X ) is assumed to follow Poisson distribution. This means the number of request in two distinct time intervals are independent from one another. Also the probability of success is linear within a time interval.

- The time interval is basically the time required for a poisson event to occur. Consequently, each distributions is defined by its parameter(s).

- Poisson distribution is defined by " Rate at which the event occurs " - ( λ ). So in our case the rate at which a support request arrives in a defined time interval. We define our distributions as follows:

                                   X ~ Po ( λ )

                                 

Where,                        λ = 1 / 30 mins

Hence,

                                   X ~ Po ( 1/30 )

a)

- We see that the time interval for events has been expanded from 30 minutes to 1 hour. However, the rate ( λ ) is given per 30 mins. In such cases we utilize the second property of Poisson distribution i.e the probability of occurrence is proportional within a time interval. Then we scale the given rate to a larger time interval as follows:

                                   λ* =  \frac{1}{\frac{1}{2} hr} = \frac{2}{1hr}

- We redefine our distribution as follows:

                                   X ~ Po ( 2/1 hr )

- Next we utilize the probability density function for poisson process and accumulate the probability for 2 to 4 request in an hour.

                           P ( X = x ) = \frac{e^-^l^a^m^b^d^a . lambda^x}{x!}

- The required probability is:

                   P ( 2 \leq X \leq 4 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )\\\\P ( 2 \leq X \leq 4 ) = \frac{e^-^2 . 2^2}{2!} +  \frac{e^-^2 . 2^3}{3!} + \frac{e^-^2 . 2^4}{4!}\\\\P ( 2 \leq X \leq 4 ) = 0.27067 + 0.18044 + 0.09022\\\\P ( 2 \leq X \leq 4 ) = 0.5413            Answer    

b)

We will repeat the process we did in the previous part and scale the poisson parameter ( λ ) to a 10 hour work interval as follows:

                               λ* = \frac{2}{1 hr} * \frac{10}{10} = \frac{20}{10 hr}

- The expected value of the poisson distribution is given as:

                             E ( X ) = λ

Hence,

                            E ( X ) = 20  (10 hour work day)    .... Answer

c)

- We redefine our distribution as follows:

                                   X ~ Po ( 20/10 hr )

- Next we utilize the probability density function for poisson process and accumulate the probability for 20 to 24 request in an 10 hour work day.

                           P ( X = x ) = \frac{e^-^l^a^m^b^d^a . lambda^x}{x!}

- The required probability is:

                   P ( 20 \leq X \leq 24 ) = P ( X = 20 ) + P ( X = 21 ) + P ( X = 22 )+P ( X = 23 ) + P ( X = 24 )\\\\P ( 20 \leq X \leq 24 ) = \frac{e^-^2^0 . 20^2^0}{20!} +  \frac{e^-^2^0 . 20^2^1}{21!} + \frac{e^-^2^0 . 20^2^2}{22!} + \frac{e^-^2^0 . 20^2^3}{23!} + \frac{e^-^2^0 . 20^2^4}{24!} \\\\P ( 20 \leq X \leq 24 ) = 0.0883 +0.08460 +0.07691 +0.06688+0.05573\\\\P ( 20 \leq X \leq 24 ) = 0.3724            Answer  

c)

The standard deviation of the poisson process is determined from the application of Poisson Limit theorem. I.e Normal approximation of Poisson distribution. The results are:

                                σ = √λ

                                σ = √20

                                σ = 4.4721 ... Answer

8 0
3 years ago
Find the cost of twelve $1,000 bonds, at a quotient of 84.50
Natali [406]
Wouldn't you have to multiply 12 time 1,000 then that by 84.50? Hope this helps
8 0
3 years ago
If it takes 5 gallons of gas to drive 115 miles, how many miles can be driven using 7 gallons of gas?
elena-s [515]

Answer: The answer is 161

Step-by-step explanation: First you write 5 and 115 as a proportion, 5/115, (for every 5 gallons you can go 115 miles). Then you find out how much you can go with 1 gallon, divide by 5. Then multiply that by 7, since that is how many gallons you need to find out.

Hope this is helpful!

7 0
3 years ago
If it is now 5:00 AM, what time will it be in 1000 hours
VikaD [51]

Answer:

9:00pm

Step-by-step explanation:

If now is 5:00 am

After 1000hrs

24hrs makes a day

1000hrs = 1000/24 = 41 days remainder 16hrs

Therefore 5:00 + 16 hrs = 2100 hr

= 9:00pm

I hope this was helpful, please mark as brainliest

5 0
3 years ago
How do you find percent of a number mentally
olga55 [171]
It isn't easy really but you can always divide the number out. For example 50% of 200 is 100 by dividing 200 by 2 or 25% of 100 is 25 by dividing 100 by 4 since 25% is equivalent to 1/4
5 0
3 years ago
Read 2 more answers
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