the j.o supplies company buys calculators from a supplier. The probability of a defective calculator is 10%. If 100 calculators are selected at random, what is the standard deviation of the number of defectives.

Standard deviation = √npq

n = 100
p = .1
q = 1 - p

I'll let you take it from here.

To find the standard deviation of the number of defectives, we need to use the binomial distribution formula:

Standard Deviation (σ) = √(nP(1-P))

Where:
n = number of trials
P = probability of success

In this case,
n = 100 (number of calculators selected)
P = 0.10 (probability of a defective calculator)

Substituting these values into the formula, we get:

σ = √(100 * 0.10 * (1 - 0.10))
= √(100 * 0.10 * 0.90)
= √(9)
= 3

Therefore, the standard deviation of the number of defectives is 3.

To calculate the standard deviation of the number of defectives, we need to first determine the mean (expected value) and variance of the number of defectives.

The mean (expected value) can be calculated by multiplying the probability of a defective calculator (p) by the total number of calculators (n). In this case, p = 0.10 (10%) and n = 100:

Mean = p * n = 0.10 * 100 = 10

The variance can be calculated by multiplying the probability of a defective calculator (p) by the probability of a non-defective calculator (q = 1 - p), and then multiplying that result by the total number of calculators (n):

Variance = p * q * n = 0.10 * (1 - 0.10) * 100 = 0.09 * 100 = 9

Finally, to obtain the standard deviation, we take the square root of the variance:

Standard Deviation = √Variance = √9 = 3

Therefore, the standard deviation of the number of defectives among the 100 selected calculators is 3.