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fenix001 [56]
2 years ago
6

A textbook has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly locat

ed on those pages. Find the probability that a random selection of 50 pages will contain no errors. Find the probability that 50 randomly selected pages will contain at least two errors.
Mathematics
1 answer:
DiKsa [7]2 years ago
6 0

Answer:

The probability of a  selection of 50 pages will contain no errors  is  0.368

The probability that the selection of the random pages will contain at least two errors is 0.2644

Step-by-step explanation:

From the information given:

Let q represent the no of typographical errors.

Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let \mu be the random variable that follows a Poisson distribution, then mean \mu = \dfrac{10}{500}= 0.02

and the mean that the random selection of 50 pages will contain no error is \lambda = 50 \times 0.02 =1

∴

Pr(q= 0) = \dfrac{e^{-1} (1)^0}{0!}

Pr(q =0) = 0.368

The probability of a  selection of 50 pages will contain no errors  is  0.368

The probability that 50 randomly page contains at least 2 errors is computed as follows:

P(X ≥ 2) = 1 - P( X < 2)

P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )]    since it is less than 2

P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]

P(X \geq 2) = 1 - [0.3678 +0.3678]

P(X \geq 2) = 1 -0.7356

P(X ≥ 2) = 0.2644

The probability that the selection of the random pages will contain at least two errors is 0.2644

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The monthly issues of the Journal of Finance are available on the Internet. The table below shows the number of times an issue w
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Answer:

Kindly check explanation

Step-by-step explanation:

Given the data below :

312 2753 2595 6057 7624 6624 6362 6575 7760 7085 7272 5967 5256 6160 6238 6709 7193 5631 6490 6682 7829 7091 6871 6230 7253 5507 5676 6974 6915 4999 5689 6143 7086

The maximum value = 7829

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Range (maximum - minimum) = (7829 - 312) = 7517

Mode(most frequently occurring) = all observations occur once

Median = 1/2 (n+1)th term

= 1/2 (33+1) = 1/2 (34) = 17th term = 6490 (after rearranging in ascending order)

Mean(m) = Σ(X) /number of observations (n) = (201608)/33 = 6109.33

Variance = Σ(x - m)² / n = 78234131 / 33 = 2370731.2

Standard deviation = sqrt(variance) = sqrt(2370731.2) = 1539.7179

5 0
2 years ago
Eric plays basketball and volleyball for a total of 95 minutes every day. He plays basketball for 25 minutes longer than he play
ioda

Answer:

Part A:

x+y= 95

x = y+25

Part B : 35 minutes

Part C : No

Step-by-step explanation:

  • Part A :

Eric plays basketball and volleyball for a total of 95 minutes every day

x+y= 95

Where:

x =the number of minutes Eric plays basketball

y= the number of minutes he plays volleyball  

He plays basketball for 25 minutes longer than he plays volleyball.

x = y+25

System:

x+y= 95

x = y+25

  • Part B:

Replacing x=y+25 on the first equation:

(y+25) + y =95

Solving for Y

y+25+y =95

25+2y=95

2y=95-25

2y=70

y = 70/2

y = 35 minutes

Part C : No

if x = 35

x+y= 95

35+y =95

y= 95-35

y = 60 minutes

Replacing y=60 on the other equation:

x = y+25

35 = 60+25

35 ≠85

5 0
3 years ago
In a basketball tournament, Team A scored 5 fewer than twice as many points as Team B. Team C scored 80 more points than Team B.
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Answer:

(2b - 5) + b + (b + 80) = 983

Step-by-step explanation:

Given that,

Total score of three teams = 983

Since the teams' scores are given in reference to team B's score, let 'b' be the score of team B.

So,

The scores of Team A = (2 * b) - 5

The scores of Team B = b

The scores of Team C = (b + 80)

Thus,

The equation for determining the total points of Team B would be:

(2b - 5) + b + (b + 80) = 983

On solving,

(2b - 5) + b + (b + 80) = 983

⇒ 2b + b + b = 983 + 5 - 80

⇒ 4b = 908

⇒ b = 908 ÷ 4

⇒ b = 227

Team B's score = 227

Team A's score = (2 * 227 - 5)

= 449

Team C's score = 227 + 80

= 307

Total ⇒ 449 + 227 + 307 = 983

Hence proved.

6 0
2 years ago
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