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iren2701 [21]
3 years ago
14

Hurry a lot of Points!

Mathematics
2 answers:
yulyashka [42]3 years ago
5 0

Answer:

17/20

Step-by-step explanation:

not even that much points but ok lol

BartSMP [9]3 years ago
5 0

Answer:

17/20

Step-by-step explanation:

points +brainliest

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A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude
EleoNora [17]

Answer:

a) P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

b) P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

c) P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

d) E(X) = 20*0.2= 4

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem  

Let X the random variable of interest, on this case we now that:  

X \sim Binom(n=20, p=0.2)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Part a

We want this probability:

P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

Part b

We want this probability:

P(X=4)

And using the probability mass function we got:

P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

Part c

We want this probability:

P(X>3)

We can use the complement rule and we got:

P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

Part d

The expected value is given by:

E(X) = np

And replacing we got:

E(X) = 20*0.2= 4

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Let x be the number. x= 2x-17
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7 • 5



Multiplication is basically repeated addition. So to convert repeated addition to multiplication, we take the number being added (7) and multiply it be the number of addens (5). Therefore, 7+7+7+7+7 = 5 • 7
3 0
3 years ago
2.4n + 1.2n - 0.12 = 7.08
Basile [38]

Answer:

n=2

Step-by-step explanation:

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3 years ago
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What is 5 2/3 as a improper fraction
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17/3. 5 times 3 is 15. Add the 2 from the numerator to get 17/3

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