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Andrew [12]
3 years ago
15

Five percent of all vcrs manufactured by a large electronics company are defective. a quality control inspector randomly inspect

s groups of 10 vcrs from the production line. using the tables for the binomial probability distribution, what is the probability that exactly one of the vcrs is defective in any one group of 10 vcrs?
Mathematics
1 answer:
harkovskaia [24]3 years ago
7 0
Binomial
Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n=10)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (defective/normal)3. Probability is known and remains constant throughout the trials (p=5%)4. All trials are random and independent of the others (assumed from context)The number of successes, x, is then given byP(x)=C(n,x)p^x(1-p)^{n-x}whereC(n,x)=\frac{n!}{x!(n-x)!}
Substituting values, p=0.05, n=10, X=exactly 1
for X=1 (defective out of n)
P(X=1)=C(10,1)0.05^1*(1-0.05)^(10-1)
=10!/(1!9!)*0.05*0.95^9
=10*0.05*0.0630249
=0.315125 (to 6 places of decimal)
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Answer:

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Step-by-step explanation:

3x-2>5x+10

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Divide the numbers: \frac{12}{2} =6

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3 years ago
the area of Alaska is about 2 x 10 to the 6 power square kilometers the area of Rhode Island is about 4 x 10 to the 3rd power sq
hram777 [196]

Answer:

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Step-by-step explanation:

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. Lightning Strikes It has been said that the probability of being struck by lightning is about 1 in 750,000, but under what cir
mrs_skeptik [129]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

      P(U |D ) = 0.198

b

   P(O\ n \ B) = 0.188

c

  P(O | B) =   0.498

Step-by-step explanation:

The total number of deaths is mathematically represented as

      T   =  16 +  23 + \cdots  +  16

        T   =623

The total number of deaths in 1996 - 2000 is mathematically represented as

     T_a =  16 +  23+ \cdots + 30

      T_a = 235

The total number of deaths in 2001 - 2005 is mathematically represented as

     T_b =  17 +  16 + \cdots + 23

      T_b = 206

The total number of deaths in 2006 - 2010 is mathematically represented as

     T_c =  15 +  17 + \cdots + 16

      T_d = 182

Generally the the probability that it would occur under the tree given that the death was  after  2000 is mathematically represented as

     P(U |D ) = \frac{P(A \ n\  U )}{P(A)}

Here  P(A \ n\  U ) represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

         P(A \ n\  U )   = \frac{Z}{ T}

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

         Z =  35 +  42

=>       Z =  77

=>       P(A \ n\  U )   = \frac{77}{ 623}

=>      

Also

     P(A) is the probability of the death occurring after 2000  and this is mathematically represented as

        P(A) =  \frac{T_b  +  T_c}{ T}

=>      P(A) =  \frac{ 206+  182}{623}

=>  

So

         P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+  182}{623}}

=>      P(U |D ) = 0.198

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

      P(O | B) = \frac{T_z}{ T}

Here T_z is the total number of death outside / camping before 2001  and the value is  117  

So

            P(O \ n \ B) = \frac{117}{623}

=>          P(O\ n \ B) = 0.188

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

       P(O | B) =  \frac{ P( O \ n \ B)}{ P(B)}

Here P(B) is the probability that it was before 2001 , this is mathematically represented as  

          P(B ) =  \frac{T_a}{T}

=>       P(B ) =  \frac{235}{623}

So

          P(O | B) =  \frac{ \frac{117}{623}}{ \frac{235}{623}}

=>       P(O | B) =   0.498

5 0
3 years ago
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