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

false

step-by-step explanation:

sas (side angle side)

aa (angle angle)

sss(side side side)

sas find two sides with an angle in between

aa is finding two angles and the third will add up to 180

sss find all sides

step-by-step explanation: