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andreev551 [17]
3 years ago
14

A farmer who grows genetically engineered corn is experiencing trouble with corn borers. A random check of 5,000 ears revealed t

he following: many of the ears contained no borers. Some ears had one borer; a few had two borers; and so on. The distribution of the number of borers per ear approximated the Poisson distribution. The farmer counted 3,500 borers in the 5,000 ears. What is the probability that an ear of corn selected at random will contain no borers? 0.3476 0.4966 1.000 0.0631
Mathematics
1 answer:
Ivenika [448]3 years ago
6 0

Answer:

Probability that an ear of corn selected at random will contain no borers is 0.4966.

Step-by-step explanation:

We are given that the distribution of the number of borers per ear approximated the Poisson distribution. The farmer counted 3,500 borers in the 5,000 ears.

Let X = <u><em>Number of borers per ear</em></u>

The probability distribution of the Poisson distribution is given by;

P(X=x) = \frac{e^{-\lambda }\times \lambda^{x}  }{x!}  ; x = 0,1,2,3,......

where, \lambda = parameter of this distribution and in our question it is proportion of bores in the total ears =  \frac{3500}{5000}  = 0.7

SO, X ~ Poisson(\lambda = 0.7)

Now, probability that an ear of corn selected at random will contain no borers is given by = P(X = 0)

                            P(X = 0)  =  \frac{e^{-0.7}\times 0.7^{0}  }{0!}

                                           =  e^{-0.7} = <u>0.4966</u>

Hence, the required probability is 0.4966.

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The following statistics represent weekly salaries at a construction company.
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Answer:

The most common salary is $605. The salary that half the employees salaries surpass is $630. The percent of employees salary that serve because $700 is 75%. Percent of employees salaries that were less than $480 is 25%. The 9% percent of employees salaries this surpass $891 years. The total weekly salary of 104 employees is 57200.

Step-by-step explanation:

It is given that mean is $550, first quartile is $480,  median is $630, third quartile is $700,  mode is $605 and 91st percentile is $891.

First quartile is at 25% of the data, median is at 50% of the data, third quartile is at 75% of the data.

The mode of a set of data values is the value that occurs most often.

The most common salary is mode, therefore the most common salary is $605.

The salary that half the employees salaries surpass is $630. Because median is the half of the data.

The percent of employees salary that serve because $700 is 75%. Because $700 is third quartile.

Percent of employees salaries that were less than $480 is 25%. Because $480 is first quartile.

The percent of employees salaries this surpass $891 years is 9%. Because $891 is 91 percentile. Therefore 9% employees get more than $891.

If the company has 104 employees than the total salary of employees is

104\times 550=57200

because means is $550, therefore the total weekly salary of 104 employees is 57200.

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<u>Step-by-step explanation:</u>

see attached graph.

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Horizontal asymptote is: y = 0  so the range cannot include 0


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