Americas
Europe
Q104E
Expert-verifiedPrivacy and information sharing. Refer to the Pew Internet & American Life Project Survey (January 2016), Exercise 4.48 (p. 239). The survey revealed that half of all U.S. adults would agree to participate in a free cost-saving loyalty card program at a grocery store, even if the store could potentially sell these data on the customer’s shopping habits to third parties. In a random sample of 250 U.S. adults, let x be the number who would participate in the free loyalty card program.
a. Find the mean of x. (This value should agree with your answer to Exercise 4.48c.)
b. Find the standard deviation of x.
c. Find the z-score for the value x = 200.
d. Find the approximate probability that the number of the 250 adults who would participate in the free loyalty card program is less than or equal to 200.
$a.\mu =125\phantom{\rule{0ex}{0ex}}b.\sigma =7.9057\phantom{\rule{0ex}{0ex}}c.z=9.4868\phantom{\rule{0ex}{0ex}}d.\mathrm{The}\mathrm{approximate}\mathrm{probability}\mathrm{is}1$
Referring to the Paw Internet & American Life Project Survey (January 2016), exercise 4.48.
x be the participate in the free loyalty card program, which follows a binomial distribution with n = 250 and p = 0.5
$a.\phantom{\rule{0ex}{0ex}}\mu =np\phantom{\rule{0ex}{0ex}}=250\times 0.5\phantom{\rule{0ex}{0ex}}=125\phantom{\rule{0ex}{0ex}}\mu =125$
Therefore, the mean of x is 125.
$b.\phantom{\rule{0ex}{0ex}}\sigma =\sqrt{np(1-p)}\phantom{\rule{0ex}{0ex}}=\sqrt{250\times 0.5\times (1-0.5)}\phantom{\rule{0ex}{0ex}}=\sqrt{250\times 0.5\times 0.5}\phantom{\rule{0ex}{0ex}}=7.9057\phantom{\rule{0ex}{0ex}}\sigma =7.9057$
Therefore, the variance of x is 7.9057.
$c.\phantom{\rule{0ex}{0ex}}\mu =125\phantom{\rule{0ex}{0ex}}\sigma =7.9057\phantom{\rule{0ex}{0ex}}x=200$
The z-score is,
$z=\frac{x-\mu}{\sigma}\phantom{\rule{0ex}{0ex}}=\frac{200-125}{7.9057}\phantom{\rule{0ex}{0ex}}=9.4868\phantom{\rule{0ex}{0ex}}z=9.4868$
Therefore, the z-score is 9.4868.
$d.\phantom{\rule{0ex}{0ex}}P(x\le 200)=P(x\le 200)\phantom{\rule{0ex}{0ex}}=P(z\le 9.4868)\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}P(x\le 200)=1$
Therefore, the approximate probability is 1.
Stock market participation and IQ. Refer to The Journal of Finance (December 2011) study of whether the decision to invest in the stock market is dependent on IQ, Exercise 3.46 (p. 182). Recall that an IQ score (from a low score of 1 to a high score of 9) was determined for each in a sample of 158,044 Finnish citizens. Also recorded was whether or not the citizen invested in the stock market. The accompanying table gives the number of Finnish citizens in each IQ score/investment category. Which group of Finnish citizens (market investors or noninvestors) has the highest average IQ score?
IQ Score | Invest in market | No investment | Totals |
1 | 893 | 4659 | 5552 |
2 | 1340 | 9409 | 10749 |
3 | 2009 | 9993 | 12002 |
4 | 5358 | 19682 | 25040 |
5 | 8484 | 24640 | 33124 |
6 | 10270 | 21673 | 31943 |
7 | 6698 | 11260 | 17958 |
8 | 5135 | 7010 | 12145 |
9 | 4464 | 5067 | 9531 |
Totals | 44651 | 113393 | 158044 |
94% of StudySmarter users get better grades.
Sign up for free