In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidence interval for the population proportion, p?

-0.0081<p<0.4081

Step-by-step explanation:

Using the formula for calculating the confidence interval expressed as;

CI = p±z√p(1-p)/n

p is the proportion = x/n

n is the sample size

x is the amount that are left-handed

p = 2/10

p = 0.2

z is the z score at 90% CI = 1.645

Substitute the values into the formula;

CI = 0.2±1.645√0.2(1-0.2)/10

CI = 0.2±1.645√0.2(0.8)/10

CI = 0.2±1.645√0.16/10

CI = 0.2±1.645√0.016

CI = 0.2±(1.645*0.1264)

CI = 0.2±0.2081

CI = (0.2-0.2081, 0.2+0.2081)

CI = (-0.0081, 0.4081)

hence the 90% confidence interval for the population proportion, p is expressed as -0.0081<p<0.4081

c.

step-by-step explanation:

it would have to be 3/0