In: Statistics and Probability
What is the probability that in the sample between 33% and 48% say that having a flexible work schedule is either very important or extremely important to their career success?
What is the probability that in the sample, between 23% and 31% are more likely to buy stock in a company based in Country A, or shop at its stores, if it is making an effort to publicly talk about how it is becoming more sustainable?
1
a.
16.5 < x < 17.5
z = (x-mean)/sd
(16.5-17)/7 < z < (17.5-17)/7
-1/14 < z < 1/14
P(-0.07 < z < 0.07)
= P(z < 0.07) - P(z < -0.07)
= 0.5279 - 0.4721 {from table below}
P(-0.07 < z < 0.07) = 0.0558
P(16.5 < x < 17.5) = 0.0558
2.
n = 100
p = 0.41
1-p = 0.59
we want x between 33 and 48
P(33 <= x <= 48) = P(33) + P(34) + ..... + P(48)
P(33 <= x <= 48) = 0.895
3.
n = 100
p = 0.27
1-p = 0.73
we want x between 23 and 31
P(23 <= x <= 31) = P(23) + P(24) + ..... + P(31)
P(23 <= x <= 31) = 0.6895
P.S. (please upvote if you find the answer satisfactory)