Question

We often use sampling in conjunction with deciding the quality of entire shipments. Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take a random sample of 25. When sampling with replacement (so that the p = probability of success does not change), note that a success in this case is selecting a defective part.

1. The mean of this situation is?

1. 12.500
2. 8.173
3. 2.500
4. 4.250

2. The P(X<6)=?

         a. 0.242

         b. 0.182

         c. 0.758

         d. 0.600

3. The P(X≤4)=?

         a. 0.622

         b. 0.750

         c. 0.424

         d. 0.576

4. The Standard deviation of this situation is?

         a. 3.767

         b. 2.811

         c. 1.878

         d. 2.294

Answer 1

Solution :

Given that,

p = 68 / 400 = 0.17

q = 1 - p = 1 - 0.17 = 0.83

n = 25

Using binomial distribution,

1)

mean = n * p = 25 * 0.17 = 4.25

2)

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

=  ((25! / 0! (25)!) * 0.17⁰ * (0.83)²⁵ +  ((25! / 1! (24)!) * 0.17¹ * (0.83)²⁴ +  ((25! / 2! (35)!) * 0.17² * (0.83)²³

+  ((25! / 3! (22)!) * 0.17³ * (0.83)²² +  ((25! / 4! (21)!) * 0.17⁴ * (0.83)²¹ +  ((25! / 5! (20)!) * 0.17⁵ * (0.83)²⁰

= 0.758

P(X < 6) = 0.758

3)

P(X 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X 4) = 0.576

4)

standard deviation = n * p * q

= 25 * 0.17 * 0.83

= 3.5275

= 1.878

standard deviation = 1.878