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Monica [59]
3 years ago
14

Each of 12 refrigerators of a certain type have been returned to a distributor because of the presence of a high pitched oscilla

ting noise when the fridge is running. Suppose that 5 of these 12 have defective compressors and the other 7 have less serious problems. If they are examined in a random order, let X= the number among the first 6 examined that have a defective compressor. Compute the following
A) P(X=5) _____

B) The probability that X exceeds its mean by more than 1 standard deviation _____
Mathematics
1 answer:
Bond [772]3 years ago
3 0

Answer:

a) P(X = 5) = 0.04397

b) The probability that X exceeds its mean by more than 1 standard deviation = P(z > 1) = 0.159

Step-by-step explanation:

a) This question can be solved using binomial distribution function formula

Using Binomial,

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of refrigerators to be examined before getting 5 refrigerators with defective compressors = 6

x = Number of successes required = 5

p = probability of success = probability of examining a refrigerator with defective compressor out of the total 12 = 5/12 = 0.4167

q = probability of failure = probability of examining a refrigerator without defective compressor out of the total 12 = 1 - (5/12) = 7/12 = 0.5833

P(X = 5) = ⁶C₅ (0.4167)⁵ (0.5833)⁶⁻⁵

P(X = 5) = ⁶C₅ (0.4167)⁵ (0.5833) = 1(0.01256)(0.5833) = 0.04397

P(X = 5) = 0.04397

b) The probability that X exceeds its mean by more than 1 standard deviation represents Z-score of z > 1

Using the normal distribution tables,

P(z > 1) = 1 - P(z ≤ 1) = 1 - 0.841 = 0.159

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Answer:

4.4722

Step-by-step explanation:

The following table shows the 36 elements of the sample space of the experiment

\left[\begin{array}{cccccccc}&1&2&3&4&5&6\\--&--&--&--&--&--&--\\1\;|&1&2&3&4&5&6\\2\;|&2&2&3&4&5&6\\3\;|&3&3&3&4&5&6\\4\;|&4&4&4&4&5&6\\5\;|&5&5&5&5&5&6\\6\;|&6&6&6&6&6&6\end{array}\right]

From this table we can compute the probabilities :

P(X = 1) = 1/36

P(X = 2) = 3/36

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P(X = 4) = 7/36

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So the expected value equals

1P(X=1)+2P(X=2)+3P(X=3)+4P(X=4)+5P(X=5)+6P(X=6) =

= 1/36 + 6/36 + 15/36 + 28/36 + 45/36 + 66/36 = 161/36 =  

4.4722

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