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aksik [14]
3 years ago
12

Each item produced by a certain manufacturer is independently of acceptable quality with probability 0.95. Approximate the proba

bility that at most 10 of the next 150 items produced are unacceptable.
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

Step-by-step explanation:

Let <em>X</em> = number of items with unacceptable quality.

The probability of an item being unacceptable is, P (X) = <em>p</em> = 0.05.

The sample of items selected is of size, <em>n</em> = 150.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and <em>p</em> = 0.05.

According to the Central limit theorem, if a sample of large size (<em>n</em> > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.

The mean of this sampling distribution is: \mu_{\hat p}= p=0.05

The standard deviation of this sampling distribution is: \sigma_{\hat p}=\sqrt{\frac{ p(1-p)}{n}}=\sqrt{\frac{0.05(1-.0.05)}{150} }=0.0178

If 10 of the 150 items produced are unacceptable then the probability of this event is:

\hat p=\frac{10}{150}=0.067

Compute the value of P(\hat p\leq 0.067) as follows:

P(\hat p\leq 0.067)=P(\frac{\hat p-\mu_{p}}{\sigma_{p}} \leq\frac{0.067-0.05}{0.0178})=P(Z\leq 0.96)=0.8315

*Use a <em>z</em>-table for the probability.

Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.

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Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen
SSSSS [86.1K]

Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ =  \frac{4!}{1! (4 - 1 )!} × \frac{6!}{2! (6 - 2 )!}

                                                                            = \frac{4!}{(3)!} × \frac{6!}{2! (4)!}

                                                                            = \frac{4.3!}{(3)!} × \frac{6.5.4!}{2! (4)!}

                                                                            = 4 × \frac{6.5}{2.1! }

                                                                            = 4 × 15 = 60

Total Number of possibility = ¹⁰C₃ = \frac{10!}{3! (10-3)!}

                                                        = \frac{10!}{3! (7)!}

                                                        = \frac{10.9.8.7!}{3! (7)!}

                                                        = \frac{10.9.8}{3.2.1!}

                                                        = 120

So, probability = \frac{60}{120} = \frac{1}{2} = 0.5

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility  = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility   = ⁴C₀ × ⁶C₃

                   =  \frac{4!}{0! (4 - 0)!} × \frac{6!}{3! (6 - 3)!}

                   = \frac{4!}{(4)!} × \frac{6!}{3! (3)!}

                   = 1 × \frac{6.5.4.3!}{3.2.1! (3)!}

                   = 1× \frac{6.5.4}{3.2.1! }

                   = 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = \frac{80}{120} = \frac{8}{12} = 0.66

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility  = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility   = ⁴C₂ × ⁶C₁

                   =  \frac{4!}{2! (4 - 2)!} × \frac{6!}{1! (6 - 1)!}

                   = \frac{4!}{2! (2)!} × \frac{6!}{1! (5)!}

                   = \frac{4.3.2!}{2! (2)!} × \frac{6.5!}{1! (5)!}

                   = \frac{4.3}{2.1!} × \frac{6}{1}

                   = 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility   = ⁴C₃ × ⁶C₀

                   =  \frac{4!}{3! (4 - 3)!} × \frac{6!}{0! (6 - 0)!}

                   = \frac{4!}{3! (1)!} × \frac{6!}{(6)!}

                   = \frac{4.3!}{3!} × 1

                   = 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = \frac{100}{120} = \frac{10}{12} = 0.83

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2 years ago
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I'm not sure what this means. If you have choices you should list them.

(1/2)*(1/4 + 1/6) is an example of what should be given. There are two ways to solve this.

1. Use the distributive property.

1/2*1/4 + 1/2* 1/6

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3/24 + 2/24 = 5/24

Method 2

Add what is inside the brackets first.

1/2 ( 1/4 + 1/6)

1/2(3/12 + 2/12 = 5/12

Now multiply by 1/2

1/2(5/12) = 5*1/(12 * 2) = 5 / 24 Same answer.

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