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IgorLugansk [536]
3 years ago
8

What is the distance from A to A’?

Mathematics
1 answer:
sergey [27]3 years ago
3 0

Answer: I think problem 2

Step-by-step explanation: hope this helps

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3 years ago
Distribution with a mean of 100 and standard deviation of 15...
anzhelika [568]
Use the z-score:
                                      112 - 100
For 112, the z-score is --------------- = 12/15, or 0.8.  
                                             15

Next, find the AREA under the standard normal curve TO THE LEFT OF 0.8.

Use either a z-score table or a calculator for this purpose.

Using my calculator, I found that this area is 0.532.  That places 112 in the 53rd percentile.

82 would be in a lower percentile:  the 12th percentile.

Let me know if you have questions about how to use a calculator to do these calculations.

8 0
4 years ago
Each item produced by a certain manufacturer is independently of acceptable quality with probability 0.95. Approximate the proba
Diano4ka-milaya [45]

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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3 years ago
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