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Levart [38]
3 years ago
13

Before a chair manufacturer sells its beanbag chairs, they spot check a random sample of chairs on the production line. The tabl

e below shows the number of common problems found during one such spot check. Common Problems Frequency Open seam 4 Cuts in upholstery 14 Understuffed 15 None 267 Total 300 If the manufacturer makes 1500 beanbag chairs per day, how many of those chairs would they expect to be understuffed?
Mathematics
2 answers:
Colt1911 [192]3 years ago
7 0

Answer:

They would expect 75 chairs to be understuffed.

Step-by-step explanation:

fenix001 [56]3 years ago
6 0

Answer:

75

Step-by-step explanation:

Given :   In beanbags

             4 Cuts in upholstery

            15 are Understuffed  

             None 267

            Total 300

To Find : If the manufacturer makes 1500 beanbag chairs per day, how many of those chairs would they expect to be understuffed?

Solution :

We are given that out of total of 300 bags there are 15 understuffed

Let x be the no. of understuffed bean bags out of total of 1500.

⇒\frac{15}{300} =\frac{x}{1500}

⇒\frac{15*1500}{300} =x

⇒75=x

Thus  If the manufacturer makes 1500 beanbag chairs per day, 75 of those chairs would they expect to be understuffed .


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Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

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Part d: <em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

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P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}

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The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as \frac{X}{4}

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Part d:  

The probability that there is at least one arrival in a 15-second period is calculated as,

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<em>The probability of at least one arrival in a 15-second period​ is 0.9179.</em>

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