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3 years ago

Correct answers only please!

Mathematics

2 answers:

vesna_86 [32]3 years ago

7 0

Answer:

answer is a

Step-by-step explanation:

18.5+24.25+24.675+17.75=85.175

Crank3 years ago

3 0

Answer:

The correct option is A.

Step-by-step explanation:

Given information:

Homework {90, 94, 96, 90}

Class Participation {95, 99}

Test Grades {87, 75, 82 ,85}

Final Exam Grade {71}

Homework is weighted 20%, Class Participation is 25%, Test Grades are 30%, and Final Exam is 25%.

$Homework=\frac{90+94+96+90}{4}\times \frac{20}{100}=18.5$

$\text{Class Participation}=\frac{95+99}{2}\times \frac{25}{100}=24.25$

$\text{Test Grades}=\frac{87+75+82+85}{4}\times \frac{30}{100}=24.675$

$\text{Final Exam Grade}=71\times \frac{25}{100}=17.25$

Course average in AMDM is

$18.5 + 24.25 + 24.675 + 17.75=85.175\approx 85$

Therefore the correct option is A.

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2 years ago

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2 years ago

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Step-by-step explanation:

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2 years ago

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2 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago