At an outpatient mental health clinic, appointment cancellations occur at a mean

Question

At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.1 per day on a typical Wednesday. Let X be the number of cancellations on a particular Wednesday.

What is the probability that four or more cancellations will occur on a particular wednesday? (Round your answer to 4 decimal places.)

The probability of a manufacturing defect in an aluminum beverage can is .00003. If 100,100 cans are produced, find the following probabilities.

At least two defective cans. (Round your answer to 4 decimal places.)

At least three defective cans. (Round your answer to 4 decimal places.)

At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.1 per day on a typical Wednesday. Let X be the number of cancellations on a particular Wednesday.

(a)

What is the probability that four or more cancellations will occur on a particular wednesday? (Round your answer to 4 decimal places.)

The probability of a manufacturing defect in an aluminum beverage can is .00003. If 100,100 cans are produced, find the following probabilities.

(a)

At least two defective cans. (Round your answer to 4 decimal places.)

  Probability    (b)

At least three defective cans. (Round your answer to 4 decimal places.)

Probability   

Explanation / Answer

Result:

At an outpatient mental health clinic, appointment cancellations occur at a mean rate of 1.1 per day on a typical Wednesday. Let X be the number of cancellations on a particular Wednesday.

Poisson distribution used with mean =1.1

(a)

What is the probability that four or more cancellations will occur on a particular wednesday? (Round your answer to 4 decimal places.)

P= 0.0257

P( x 4) = 1-P( x <4)

=1-(0.3329+0.3662+0.2014+0.0738)

=0.0257

POISSON.DIST Probabilities Table

X

P(X)

0

0.3329

1

0.3662

2

0.2014

3

0.0738

The probability of a manufacturing defect in an aluminum beverage can is .00003. If 100,100 cans are produced, find the following probabilities.

Binomial distribution used, n=100100,p=0.00003

(a)

At least two defective cans. (Round your answer to 4 decimal places.)

  Probability

  

P( x 2) = 0.8013

(b)

At least three defective cans. (Round your answer to 4 decimal places.)

Probability

  

P( x 3) = 0.5775

(a)

What is the probability that four or more cancellations will occur on a particular wednesday? (Round your answer to 4 decimal places.)

P= 0.0257

P( x 4) = 1-P( x <4)

=1-(0.3329+0.3662+0.2014+0.0738)

=0.0257

POISSON.DIST Probabilities Table

X

P(X)

0

0.3329

1

0.3662

2

0.2014

3

0.0738

The probability of a manufacturing defect in an aluminum beverage can is .00003. If 100,100 cans are produced, find the following probabilities.

Binomial distribution used, n=100100,p=0.00003

(a)

At least two defective cans. (Round your answer to 4 decimal places.)

  Probability

  

P( x 2) = 0.8013

(b)

At least three defective cans. (Round your answer to 4 decimal places.)

Probability

  

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