1. A certain manufacturing process to build phone batteries results in the production of defective batteries at a rate of 5% (0.05). Suppose that 10 batteries are produced. a. Find the expected number of defective batteries. b. Find the variance and standard deviation in the number of defective batteries. c. Find the probability that at least 1 battery is defective.

0.5 ; 0.475 ; 0.689 ; 0.4013

Step-by-step explanation:

Given that:

Rate of production of defective batteries p = 0.05

Number of batteries produced (n) = 10

The expected number of defective batteries = mean = n * p = 10 * 0.05 = 0.5 batteries

Variance of defective batteries :

Var(X) = n * p * q ; q = 1 - p

Hence,

Var(X) = 10 * 0.05 * 0.95 = 0.475

Standard deviation (X) = sqrt(variance) = sqrt(0.475) = 0.689

Probability that atleast 1 battery is defective :

Using the binomial probability function

P(x ≥ 1) = 1 - p(x = 0)

= 1 - q^n

= 1 - 0.95^10

= 1 - 0.59873693923837890625

= 0.40126306076162109375

= 0.4013

boxplot

i would say b) is your answer

sorry if im wrong

step-by-step explanation: