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Vesnalui [34]
3 years ago
12

The yield of a chemical process is being studied. Fromprevious experience, yield is known to be normally distributedand 3. The p

ast five days of plant operation have resulted inthe following percent yields: 91.6, 88.75, 90.8, 89.95, and 91.3.Find a 95% two-sided confidence interval on the true mean yield.
Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

(87.85 ≤ μ ≤ 93.11)

Step-by-step explanation:

Given the data:

91.6, 88.75, 90.8, 89.95, 91.3

Mean, m = Σx / n

n = sample size = 5

Mean = 452.4 / 5 = 90.48

Standard deviation, σ = 3

Zcritical at 95% = 1.96

Confidence interval :

Mean ± Error margin

Error margin = Zcritical*σ/sqrt(n)

Error margin = 1.96 * 3/sqrt(5)

Error margin = 2.630

Lower boundary : 90.48 - 2.630 = 87.85

Upper boundary : 90.48 + 2.630 = 93.11

(87.85 ; 93.11)

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Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50
Vladimir79 [104]

Answer:

a) Var[z] = 1600

    D[z] = 40

b) Var[z] = 2304

    D[z] = 48

c) Var[z] = 80

    D[z] = 8.94

d) Var[z] = 80

    D[z] = 8.94

e) Var[z] = 320

    D[z] = 17.88

Step-by-step explanation:

In general

V([x+y] = V[x] + V[y] +2Cov[xy]

how in this problem Cov[XY] = 0, then

V[x+y] = V[x] + V[y]

Also we must use this properti of the variance  

V[ax+b] = a^{2}V[x]

and remember that

standard desviation = \sqrt{Var[x]}

a) z = 35-10x

   Var[z] = 10^{2} Var[x] = 100*16 = 1600

   D[z] = \sqrt{1600} = 40

b) z = 12x -5

   Var[z] = 12^{2} Var[x] = 144*16 = 2304

   D[z] = \sqrt{2304} = 48

c) z = x + y

   Var[z] =  Var[x+y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = \sqrt{80} = 8.94  

d) z = x - y

   Var[z] =  Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = \sqrt{80} = 8.94

e) z = -2x + 2y

   Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320

  D[z] = \sqrt{320} = 17.88

   

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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