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Ksenya-84 [330]
3 years ago
5

100 students take a course pass/fail. If they pass they get 4 points towards their GPA, if they fail they get 0. If 90 students

pass, what is the mean and standard deviation (to 1 decimal) of the points earned?
Mean: ________a0

Standard Deviation:________ a1
Mathematics
1 answer:
katrin2010 [14]3 years ago
8 0
Formulas
Mean = total sum of points / # of data
[Stantard deviation] ^ 2 = {Sum of [every data - mean]^2 } / [number of data - 1]

Procedure:

Mean = total sum of points / # of data

Total sum of points = point of pass + point of fail

points of pass = 90*4 = 360
point of fail = 10*0 = 0
Total sum of points = 360

Number of data = 100

Mean = 360/100 = 3.6

[Stantard deviation] ^ 2 = {Sum of [every data - mean]^2 } / [number of data - 1]

[Sum of every data - mean]^2  = 90*[4 -3.6]^2 + 10* [0 - 3.6]^2 = 14.4 + 129.6 = 144

 [Stantard deviation] ^ 2 = [144]/[100-1] =144/99

Standar deviation = √(144/99) = 1.206


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2. The quality assurance department inspects its production line. The product either fails or passes the inspection. Past experi
Oksana_A [137]

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probability of success = p = 1 - 0.05 = 0.95

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(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)} $

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\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} )\\\\

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

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Given:

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Let Mrs. Charles owns x part of business.

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Therefore, Mrs. Brown owns  \dfrac{12}{19} of the business.

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