Question

1. If 15% of adults in a certain country work from home, what is the probability that fewer than 42 out of a random sample of 350 adults will work from home? (Round your answer to 3 decimal places)

2. 2. Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If we want to estimate this proportion to within 4% at the 95% confidence level, how many randomly selected teenagers must we survey?

Please help me out these short quick questions.. Please help me out with exact and complete answers with explanations thank you very much

Answer 1

Answer)

First we need to check the conditions of normality that if n*p and n*(1-p) both are greater than 5 or not

N = 350

P = 42/350

N*(1-p) = 308

N*p = 42

As both are greater than, 5

Normality conditions are met, so we can use standard normal z table to estimate the probability

We need to find mean

P = 0.15

Mean = 350*0.15 = 52.5

Standard deviation = √n*p*(1-p)

= 6.6801946079436

Z = (x-mean)/s.d

And by uncertainty

P(x<42) = p(x<41.5)

Z = (41.5 - 52.5)/6.6801946079436 = -1.65

From z table

P(z<-1.65) = 0.0495

2)

from z table, critical value z for 95 % confidence level is 1.96

margin of error is equal to z*square root of p*(1-p)/square root of n

we need to find n

z = 1.96

best estimate of p = 0.5

margin of error is = 0.04 (4%)

after substituting all the values

n = 601