As part of their strategy to keep computer viruses in check, a company randomly

Question

As part of their strategy to keep computer viruses in check, a company randomly selects 20% of its computers to have their hard drives (HDDs) get a weekly virus scan. 50 drives exist in the marketing department of the company. What is the probability that more than 6 HDDs in marketing get scanned? What is the probability that 3 or less HDDs in marketing get scanned? On average, what is the number of HDDs that get scanned in marketing? What is the variance in the number of HDDs that get scanned weekly in marketing? What is the probability that all 20 HDDs in marketing get scanned?

Explanation / Answer

Hi! I will be using normal approximation to the binomial, as n > 30 here. If you need another approach here, please resubmit this problem indicating what approach you need. That way we can continue helping you! Thanks!

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Here, n = 50, p = 0.20.

A)

We first get the z score for the critical value:          
          
x = critical value =    6.5      
u = mean = np =    10      
          
s = standard deviation = sqrt(np(1-p)) =    2.828427125      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.237436867      
          
Thus, the left tailed area is          
          
P(z <   -1.237436867   ) =    0.892037531 [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    3.5      
u = mean = np =    10      
          
s = standard deviation = sqrt(np(1-p)) =    2.828427125      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -2.298097039      
          
Thus, the left tailed area is          
          
P(z <   -2.298097039   ) =    0.010778133 [ANSWER]

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c)

u = mean = np =    10

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d)

variance = np(1-p) = 50*0.20*(1-0.20) = 8 [ANSWER]

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E)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    19.5      
x2 = upper bound =    20.5      
u = mean = np =    10      
          
s = standard deviation = sqrt(np(1-p)) =    2.828427125      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    3.358757211      
z2 = upper z score = (x2 - u) / s =    3.712310601      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.999608531      
P(z < z2) =    0.999897312      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.000288781   [ANSWER]

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Hi! There seems to be a problem with part e), as it says "all 20 HDDs", when in fact there are 50 HDDs there. I assumed it means exactly 20 out of 50 HDDs, as it makes more sense. Please report this probelm to your instructor. Thanks!

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