Answer to Question #216175 in Statistics and Probability for Navy

Question

a file cabinet manufacturer estimates that 5% of its file cabinets will have drawers that close improperly. Assume a probably run of 40 cabinets is completed. a.what is the mean and standard deviation of this distribution? b. what is the probability that 3 or more are defective? c. what is the probability that exactly 3 cabinets are defective? d. what is the probability that there are more than 4 defective file cabinets?

Answer 1

Let $X=$ the number of defective file cabinets: $X\sim Bin(n, p).$

Given $n=40, p=0.05, q=1-p=1-0.05=0.95.$

a.

$$\mu=np=40(0.05)=2$$

$$\sigma^2=npq=40(0.05)(0.95)=1.9$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{1.9}\approx1.3784$$

b.

$$P(X\geq 3)=1-P(X=0)-P(X=1)$$

$$-P(X=2)=1-\dbinom{40}{0}(0.05)^0(0.95)^{40-0}$$

$$-\dbinom{40}{1}(0.05)^1(0.95)^{40-1}-\dbinom{40}{2}(0.05)^2(0.95)^{40-2}$$

$$\approx0.32326423925$$

c.

$$P(X= 3)=\dbinom{40}{3}(0.05)^3(0.95)^{40-3}$$

$$\approx0.18511446375$$

d.

$$P(X>4)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)-P(X=4)$$

$$=1-\dbinom{40}{0}(0.05)^0(0.95)^{40-0}$$

$$-\dbinom{40}{1}(0.05)^1(0.95)^{40-1}-\dbinom{40}{2}(0.05)^2(0.95)^{40-2}$$

$$-\dbinom{40}{3}(0.05)^3(0.95)^{40-3}-\dbinom{40}{4}(0.05)^4(0.95)^{40-4}$$

$$\approx0.04802826025$$