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andriy [413]
2 years ago
11

The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.35 inches and a standard deviatio

n of 0.01 inches. What percentage of bolts will have a diameter greater than 0.36 inches? Enter your answer as a percentage rounded to two decimal places.
Mathematics
1 answer:
Fiesta28 [93]2 years ago
7 0

Answer:

15.87%

Step-by-step explanation:

Notice that the mean of 0.35 inches with a standard deviation of 0.01 gives you when you add (to the right of the distribution), exactly 0.36. Since you want to find the probability (or percentage) of the bolts that have diameter LARGER than 0.36 in, that means you want to estimate the area under the Normal distribution curve from 0.36 to the right). See attached image.

We can use the tables of Z distribution for that, or the standard normal tables:

P(x>0.36) = P(z>(0.36-0.35)/0.01) = P(Z>1) = 0.1587 = 15.87%

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According to a study done by a university​ student, the probability a randomly selected individual will not cover his or her mou
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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 P(X = 8) =  0.0037

b

 P(X <  5) =  0.805

c

 P(X > 6) =  0.0206

I would be surprised because the value is very small , less the 0.05

Step-by-step explanation:

From the question we are told that

The probability a randomly selected individual will not cover his or her mouth when sneezing is p = 0.267

Generally data collected from this study follows  binomial  distribution because the number of trials is  finite , there are only two outcomes, (covering  , and  not covering mouth when sneezing ) , the trial are independent

Hence for a randomly selected variable  X we have that  

   X \ \ \~ \ \ { B ( p , n )}

The probability distribution function for binomial  distribution is  

    P(X = x ) =  ^nC_x *  p^x *  (1 -p) ^{n-x}

Considering question a

Generally the  the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing is mathematically represented as

     P(X = 8) =  ^{12} C_8 *  (0.267)^8 *  (1- 0.267)^{12-8}

Here C denotes  combination

So

     P(X = 8) =  495  *  0.000025828 * 0.28867947

    P(X = 8) =  0.0037

Considering question b

Generally the probability that among 12 randomly observed individuals fewer than 5 do not cover their mouth when​ sneezing is mathematically represented as

     P(X <  5 ) =[P(X = 0 ) + \cdots + P(X = 4)]

=>   P(X <  5 ) =[ ^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0} + \cdots +  ^{12} C_4 *  (0.267)^4 *  (1- 0.267)^{12-4} ]

=> P(X <  5 )  =  0.02406 +  0.10516 + 0.21067 + 0.25580 + 0.20964

=>  P(X <  5) =  0.805

Considering question c

Generally the probability that fewer than half(6) covered their mouth when​ sneezing(i.e the probability the greater than half do not cover their mouth when sneezing) is mathematically represented as

      P(X > 6) =  1 - p(X \le  6)

=>    P(X > 6) = 1 - [P(X = 0) + \cdots + P(X =6)]

=>    P(X > 6)=1 - [^{12} C_0 *  (0.267)^0 *  (1- 0.267)^{12-0}+ \cdots + ^{12} C_4 *  (0.267)^6 *  (1- 0.267)^{12-6} ]

=>    P(X > 6)= 1 - [0.02406 + \cdots + 0.0519 ]  

=>    P(X > 6) =  0.0206

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