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2 years ago

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A manager wants to build 3-sigma x-bar control limits for a process. The target value for the mean of the process is 10 units, a

nd the standard deviation of the process is 6. If samples of size 9 are to be taken, what will be the upper and lower control limits, respectively? 12 and 8 -8 and 28 4 and 16 16 and 4 8 and 12

1 answer:

lubasha [3.4K]2 years ago

3 0

Answer: Third option is correct.

Step-by-step explanation:

Since we have given that

Mean = 10

Sample size = 9

Standard deviation = 6

We need to find the upper and lower control limits.

so, Lower limit would be

$\bar{x}-3\dfrac{\sigma}{\sqrt{n}}\\\\=10-3\times \dfrac{6}{\sqrt{9}}\\\\=10-6\\\\=4$

Upper limit would be

$\bar{x}+3\dfrac{\sigma}{\sqrt{n}}\\\\=10+3\times \dfrac{6}{\sqrt{9}}\\\\=10+6\\\\=16$

Hence, Third option is correct.

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Twenty-four 4-inch wide square posts are evenly spaced with 5 feet between adjacent posts to enclose a square field, as shown. W

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Answer:

492 0/10

Step-by-step explanation:

We have that the perimeter is 4 times the length of the side, now we know that this side "l" is given by:

l = 4 in * 24 + 23 * 5 ft

l = 96 in + 115 ft

Now, we pass the inches to pes, knowing that 1 foot equals 12 inches, so

96 in * 1 ft / 12 in = 8 ft

therefore replacing it remains:

l = 8 ft + 115 ft

l = 123 ft

now the perimeter:

p = 123 * 4

p = 492

to change to a mixed number:

4920/10 = 492 0/10

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Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

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Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

4 of the next 20 drivers do you expect to come to a complete stop

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An antique store increases all of its prices by $40\%$, and then announces a $25\%$-off-everything sale. What percent of the ori

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Let X be the original price.
Do a 40% increase we get the price: X+0.4X=1.4X
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=X+0.05X.
The new prices are 5% increase from the original price.

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Answer:

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laptop: x + 3800

phone: x

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x = 10200

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phone: 10200

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Answer: 0.60

Step-by-step explanation:

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1 year ago

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