If 10% of men are bald, what is the probability that fewer than 100 in a random sample of 818 men are bald? (Answers must be in 4 decimal values)

the probability that fewer than 100 in a random sample of 818 men are bald is 0.9830

Step-by-step explanation:

Given that;

p = 10% = 0.1

so let q = 1 - p = 1 - 0.1 = 0.9

n = 818

μ = np = 818 × 0.1 = 81.8

α = √(npq) = √( 818 × 0.1 × 0.9 ) = √73.62 = 8.58

Now to find P( x < 100)

we say;

Z = (X-μ / α) = ((100-81.8) / 8.58) = 18.2 / 8.58 = 2.12

P(x<100) = P(z < 2.12)

from z-score table

P(z < 2.12) = 0.9830

Therefore the probability that fewer than 100 in a random sample of 818 men are bald is 0.9830

answer: h

step-by-step explanation:

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation: