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icang [17]
3 years ago
5

Please help me on number 10 and 11

Mathematics
1 answer:
denpristay [2]3 years ago
3 0
To solve number 10, you should do just as it says and type it into a calculator. To save you the trouble, I have already done this. It equals approximately 4.26.

As for number 11, you have not included the whole question, so there is not currently enough information to solve it. 
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The answer to the problem is 12. 2 times 3 is 6. 18 divided by 6 is 3 and 3 times 4 is 12.
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4<br>Order these decimals, starting with the smallest<br>7.96<br>9.61<br>9.534<br>9.5<br>9.53​
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7.96,9.5,9.53,9.534,9.61

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In a class of 40 students, 8 are in the drama club and 12 are in the art club. If a student is selected at random, what is the p
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total students = 40probability of selecting a student from art club= 12/40= 3/10

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2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi
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Answer:

(a) E(X) = 950

(b) $ COV = 0.007255$

(c) P(X > 980) = 0.00001\\\\

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$ COV = \frac{\sigma}{E(X)} $

Where the standard deviation is given by

\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892

So the coefficient of variance is

$ COV = \frac{6.892}{950} $

$ COV = 0.007255$

(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

P(X > 980)  = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980)  = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980)  = 1 - P(Z < 4.28)\\\\

The z-score corresponding to 4.28 is 0.99999

P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\

So it means that it is very unlikely that there will be more than 980 non-defective units.

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