Question

In: Statistics and Probability

According to a social media​ blog, time spent on a certain social networking website has a...

  1. According to a social media​ blog, time spent on a certain social networking website has a mean of 17 minutes per visit. Assume that time spent on the social networking site per visit is normally distributed and that the standard deviation is 7 minutes.
    1. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 16.5 and 17.5 minutes?
  2. A global research study found that the majority of​ today's working women would prefer a better​ work-life balance to an increased salary. One of the most important contributors to​ work-life balance identified by the survey was​ "flexibility," with 41​% of women saying that having a flexible work schedule is either very important or extremely important to their career success. Suppose you select a sample of 100 working women.

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?

  1. A survey found that 27​% of consumers from a Country A 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. Suppose you select a sample of 100 respondents from Country A. Complete parts​ (a) through​ (d) below.

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?

Solutions

Expert Solution

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)


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