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rodikova [14]
2 years ago
15

An electronic device factory is studying the length of life of the electronic components they produce. The manager takes a rando

m sample of 50 electronic components from the assembly line and records the length of life in the life test. From the sample he found the average length of life was 100,000 hours and that the standard deviation was 3,000 hours. He wants to find the confidence interval for the average length of life of the electronic components they produced. Based on the information, what advice would you give to him?
Select one or more:

a. The distribution of the length of life of the electronic components is usually right skewed. Thus, he should not compute the confidence interval.

b. He did not take a simple random sample of the electronic components; thus he should not compute the confidence interval

c. The mean and standard deviation are large enough to compute the confidence interval.

d. The population is right skewed, but the sample size is large enough to use a normal approximation. Thus he can compute the confidence interval.

e. He can calculate the confidence interval but should use a t-distribution since it deals with an average.
Mathematics
1 answer:
alisha [4.7K]2 years ago
6 0
The answer is B because if you have 10000 hours you subtract by 3000 and it’s equals b
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Answer:

See explanation

Step-by-step explanation:

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- The demand function of a certain brand is given as price P a function of x quantity of goods ( in hundred ) demanded per month. The relation is:

                           P ( x ) = 50 Ln ( 50 / x ).

- The point price elasticity ( E ) of demand is given by:

                           E = \frac{P}{x}*\frac{dP}{dx}  

- Where, dP / dx : is the rate of change of price ( P ) with each hundred unit of good ( x ) is demanded.

- To determine the " dP / dx " by taking the first derivative of the given relation:

                          P ( x ) = 50 Ln ( 50 / x ).

                          d P ( x ) / dx = [ 50*x / 50 ] * [ -1*50 / x^2 ]

                                              = - 50 / x

- Hence the point price elasticity of demand is given by:

                          E = - ( P / x ) * ( 50 / x )

                          E = -50*P / x^2    

- For an inelastic demand, ! E ! is < 1:

                          ! -50*P / x^2 ! < 1

                          50*P / x^2 < 1

                          P < x^2 / 50

- For an unitary demand, ! E ! is =  1:

                          ! -50*P / x^2 !  = 1

                          50*P / x^2 = 1

                          P = x^2 / 50

- For an inelastic demand, ! E ! is > 1:

                          ! -50*P / x^2 ! > 1

                          50*P / x^2 > 1

                          P > x^2 / 50

2)

If the unit price is increased slightly from $50, will the revenue increase or decrease?

- We see from the calculated demand sensitivity d P / dt:

                          d P ( x ) / dx = - 50 / x

- We see that as P increases the from P = $50, the quantity of goods demanded would be:

                          50 = 50 ln(50/x)

                           1 = Ln ( 50 / x )

                           50/x = e

                           x = 50 / e

Then,

                          d P ( x ) / dx = - 50 / ( 50 / e )

                          d P ( x ) / dx = - e

- We see that if price slightly increases from $ 50 then the quantity demanded would decrease by e (hundreds ) goods.

- The decrease in the quantity demanded is higher than the increase in price. The revenue is given by the product of price P ( x ) and x:

                Revenue R ( x ) = P ( x ) * x

                                = 50*x*ln(50/x)

Then the product of price and quantity goods also decreases; hence, revenue decreases.

8 0
3 years ago
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