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

could you reply to this with the question so i can ?

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."