Question

QUESTION 5 A factory produces pins of which 1.5% are defective. The components are packed in boxes of 12. A box is selected at random.

(1) n = 12

(2) p = 0.0150 (Round to four decimal places)

(3) q = (Round to four decimal places)

Find the following probabilities (Round ALL answers to four decimal places):

4) The box contains exactly 6 defective pins

5) The box contains at least one defective pins

6) The box contains no more than two defective pins

7) The box contains between four and eight defective pins

8) All the pins in the box are defective

9) The box contains less than three or more than nine defective pins

10) The box contains no defective pins

11) The box contains less than eight or between five and ten defective pins.

12) The box contains less than seven

13) The box contains more than ten

Calculate (Round ALL answers to two decimal places):

14) The mean of the variable “the number of defective pins”.

15) The standard deviation of the variable “the number of defective pins”

Please Answer all 15 question correctly

Answer 1

1) n = 12

2) p = 0.015

3) q = 1 - 0.015 = 0.985

4) P(X = 6) = 12C6 * (0.015)^6 * (0.985)^6 = 0.0000

5) P(X > 1) = 1 - P(X < 1)

                   = 1 - P(X = 0)

                   = 1 - 12C0 * (0.015)^0 * (0.985)^12

                   = 1 - 0.8341 = 0.1659

6) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

                  = 12C0 * (0.015)^0 * (0.985)^12 + 12C1 * (0.015)^1 * (0.985)^11 + 12C2 * (0.015)^2 * (0.985)^10 = 0.9993

7) P(4 < X < 8)

= P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

= 12C4 * (0.015)^4 * (0.985)^8 + 12C5 * (0.015)^5 * (0.985)^7 + 12C6 * (0.015)^6 * (0.985)^6 + 12C7 * (0.015)^7 * (0.985)^5 + 12C8 * (0.015)^8 * (0.985)^4 = 0.0000

8) P(X = 12) = 12C12 * (0.015)^12 * (0.985)^0 = 0.0000

9) P(X < 3) + P(X > 9)

   = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 10) + P(X = 11) + P(X = 12)

= 12C0 * (0.015)^0 * (0.985)^12 + 12C1 * (0.015)^1 * (0.985)^11 + 12C2 * (0.015)^2 * (0.985)^10 + 12C10 * (0.015)^10 * (0.985)^2 + 12C11 * (0.015)^11 * (0.985)^1 + 12C12 * (0.015)^12 * (0.985)^0 = 0.9993

10) P(X = 0) = 12C0 * (0.015)^0 * (0.985)^12 = 0.8341

11) P(X < 10)

= 1 - P(X > 10)

= 1 - P(X = 11) + P(X = 12)

= 1 - (12C11 * (0.015)^11 * (0.985)^1 + 12C12 * (0.015)^12 * (0.985)^0)

= 1 - 0.0000

= 1

12) P(X < 7)

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

= 0.9999

13) P(X > 10)

= P(X = 11) + P(X = 12)

= (12C11 * (0.015)^11 * (0.985)^1 + 12C12 * (0.015)^12 * (0.985)^0)

= 0.0000

14) Mean = n * p = 12 * 0.015 = 0.18

15) Standard deviation = sqrt(np(1 - p)) = sqrt(12 * 0.015 * 0.985) = 0.42