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

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

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐