Round appropriately Question 18 According to The Yankee Group, 53% of all cable
ID: 2946123 • Letter: R
Question
Round appropriatelyQuestion 18 According to The Yankee Group, 53% of all cable households rate cable companies as good or excellent in quality transmission cable companies as good or excellent in having professional personnel. Suppose 300 cable households are randomly contacted. Sixty percent of all cable houtensidiate (a) What is the probability that more than 175 cable households rate cable companies as good or excellent in quality trans mission? between 165 and 170 (inclusive) cable households rate cable companies as good or excellent in quality transi or excellent in quality transmission? households rate cable companies as good or excellent in having professional personnel? d) What is the probability that fewer than 200 cable households rate cable companies as good or excellent in having professional person (Round z values and ? values to 2 places. Round your final answers to 4 decimal places.) (a) Px > 175) (b) P(165 SxS 170) (e) P(155 s x S 170) (d) P(x
Explanation / Answer
Here we will use the Normal Approximation of the Binomial Variable where if X is Binomial Variable with parameters n and p then it has normal approximation with mean = n*p and Variance = n*p*(1-p)
So in case of Quality Transmission x will follow Normal with mean 300 * 0.53 = 159 and Variance = 300 * 0.53 *(1 - 0.53) = 74.73 . Therefore Standard Deviation is 74.730.5 = 8.645
Similarly in case of Professional Personal x will follow Normal with mean 300 * 0.6 = 180 and Variance = 300 * 0.6 * (1 - 0.6) = 72 . Therefore Standard Deviation is 720.5 = 8.485
Ans a ) P ( x > 175 ) = P ( x > 175.5) { Continuity Correction}
= P( (x - 159)/ 8.645 > (175.5 - 159)/ 8.645)
= P ( (x - 159)/ 8.645 > 1.9086)
Now (x - 159)/ 8.645 will follow Standard Normal let be denoted by z
P(z > 1.9086) = 0.0281 approximately
Ans b P( 165 <= x <= 170) = P( 164.5 < x < 170.5) (Continuity Correction)
= P( (164.5 - 159)/ 8.645 < (x - 159)/ 8.645 < (170.5 - 159)/ 8.645 )
= P( (164.5 - 159)/ 8.645 < z < (170.5 - 159)/ 8.645 )
= P(0.6362 < z < 1.3302)
= P(z < 1.3302) - P(z<0.6362) = 0.9082 - 0.7389 = 0.1693
Ans c P(155 <= x <= 170) = P( 154.5 < x < 170.5) ( Continuity Correction)
Here (x - 180) / 8.485 will follows Standard Normal incase of professional personnel
= P ( (154.5 - 180)/8.485 < (x - 180)/8.485 < (170.5 - 180)/8.485 )
= P ( -3.005 < z < -1.119)
= P ( z < -1.119) - P(z<-3.005)
= 0.1314 - 0.0013 = 0.1301
Ans d) P(x < 200) = P(x < 199.5) {Continuity Correction}
= P( (x - 180)/8.485 < (199.5 - 180)/8.485 )
= P ( z < 2.298) = 0.9893
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