12.04.2022

Suppose X has an exponential distribution with mean equal to 11. Determine the following: (a) (Round your answer to 3 decimal places.) (b) (Round your answer to 3 decimal places.) (c) (Round your answer to 3 decimal places.) (d) Find the value of x such that . (Round your answer to 2 decimal places.)

. 1

Step-by-step answer

24.06.2023, solved by verified expert
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Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Step-by-step explanation:

Given

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00 --- Mean

Required (Missing from  the question)

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

(d) x such that Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

In an exponential distribution:

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00 --- the pdf

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00 --- the cdf

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

In the above equations:

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Substitute 11 for E(x)

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Now, we solve (a) to (d) as follows:

Solving (a): P(X>11)

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Substitute 11 for x in Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Remove bracket

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Solving (b): P(X>22)

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Substitute 22 for x in Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Remove bracket

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Solving (c): P(X>33)

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Substitute 33 for x in Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Remove bracket

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Solving (d): x when Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Here, we make use of:

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Substitute Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

So, we have:

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Subtract 1 from both sides

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Reorder the equation

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Substitute 1/11 for Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Solve for x:

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00

Suppose X has an exponential distribution with, №17886720, 12.04.2022 04:00 --- approximated

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Mathematics
Step-by-step answer
P Answered by PhD

P(X  11) = 0.368

P(X  22) = 0.135

P(X  33) = 0.050

x = 33

Step-by-step explanation:

Given

E(x) = 11 --- Mean

Required (Missing from  the question)

(a)\ P(X11)

(b)\ P(X22)

(c)\ P(X33)

(d) x such that P(X

In an exponential distribution:

f(x) = \lambda e^{-\lambda x}, x \ge 0 --- the pdf

F(x) = 1 - e^{-\lambda x}, x \ge 0 --- the cdf

P(X  x) = 1 - F(x)

In the above equations:

\lambda = \frac{1}{E(x)}

Substitute 11 for E(x)

\lambda = \frac{1}{11}

Now, we solve (a) to (d) as follows:

Solving (a): P(X>11)

P(X  11) = 1 - F(11)

Substitute 11 for x in F(x) = 1 - e^{-\lambda x}

P(X  11) = 1 - (1 - e^{-\frac{1}{11}* 11})

P(X  11) = 1 - (1 - e^{-\frac{11}{11}})

P(X  11) = 1 - (1 - e^{-1})

Remove bracket

P(X  11) = 1 - 1 + e^{-1}

P(X  11) = e^{-1}

P(X  11) = 0.368

Solving (b): P(X>22)

P(X  22) = 1 - F(22)

Substitute 22 for x in F(x) = 1 - e^{-\lambda x}

P(X  22) = 1 - (1 - e^{-\frac{1}{11}* 22})

P(X  22) = 1 - (1 - e^{-\frac{22}{11}})

P(X  22) = 1 - (1 - e^{-2})

Remove bracket

P(X  22) = 1 - 1 + e^{-2}

P(X  22) = e^{-2}

P(X  22) = 0.135

Solving (c): P(X>33)

P(X  33) = 1 - F(33)

Substitute 33 for x in F(x) = 1 - e^{-\lambda x}

P(X  33) = 1 - (1 - e^{-\frac{1}{11}* 33})

P(X  33) = 1 - (1 - e^{-\frac{33}{11}})

P(X  33) = 1 - (1 - e^{-3})

Remove bracket

P(X  33) = 1 - 1 + e^{-3}

P(X  33) = e^{-3}

P(X  33) = 0.050

Solving (d): x when P(X

Here, we make use of:

P(X

Substitute F(x) = 1 - e^{-\lambda x}

P(X

So, we have:

0.95 = 1 - e^{-\lambda x}

Subtract 1 from both sides

0.95 -1= 1-1 - e^{-\lambda x}

-0.05=- e^{-\lambda x}

Reorder the equation

e^{-\lambda x} = 0.05

Substitute 1/11 for \lambda

e^{-\frac{1}{11} x} = 0.05

Solve for x:

x = -\frac{1}{1/11}\ ln(0.05)

x = -11\ ln(0.05)

x = 32.9530550091

x = 33 --- approximated

Mathematics
Step-by-step answer
P Answered by PhD
Answer: 440 grams for 1.54 is the better value
Explanation:
Take the price and divide by the number of grams
1.54 / 440 =0.0035 per gram
1.26 / 340 =0.003705882 per gram
0.0035 per gram < 0.003705882 per gram
Mathematics
Step-by-step answer
P Answered by PhD

Cost of 7 gallons=$24.50

Cost of 1 gallon=24.50/7=3.5

Cost of 15 gallons=15*3.5=52.5

Cost of 15 gallons will be $52.5

Mathematics
Step-by-step answer
P Answered by PhD

The answer is in the image 

The answer is in the image 
Mathematics
Step-by-step answer
P Answered by PhD

The answer is in the image 

The answer is in the image 
Mathematics
Step-by-step answer
P Answered by PhD

Speed=Distance/time

Here,

distance=15m

time=1sec

speed=15/1=15m/sec

Distance=Speed*time

time=15min=15*60sec=900sec

Distance travelled in 15 min=15*900=13,500m

=13500/1000 km=13.5Km

Mathematics
Step-by-step answer
P Answered by PhD

The answer is in the image 

The answer is in the image 

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