Here is the syntax you can use to check your answers. (Forward and Backward)

Say you want to know the \(Pr (\hat{P} < .35)\) and \(\hat{P} \sim \mathcal{N}(.4,.07)\)

pnorm(.35, mean = .4, sd = .07 )
## [1] 0.2375253

Here is the syntax you can use to check if a “Backward” calcuation is corect.

Say you know the probability to the left of \(\hat{p}\) = .04 and you want to know what the appropriate \(\hat{p}\) is. You also know that \(\hat{P} \sim \mathcal{N}(.4,.07)\)

qnorm(.05, mean = 12, sd = 4)
## [1] 5.420585

Section 8.2

11.

  1. Sample of n=75 is less than 5% of population, and np(1-p) is 12, so the sampling distribution is normal.
  2. z = .84-.8/(.0462) = .87; convert to normal probability = .8078; probability = 0.1922
  3. z = .68-.8/(.0462) = -2.59; convert to normal probability = .0048

12.

  1. Sample of n=200 is less than 5% of population, and np(1-p) is greater than 10 (45.5), so the sampling distribution is normal.
  2. z = .68-.65/(.0337) = .89; convert to normal probability = .8133; probability = 0.1867
  3. z = .59-.65/(.0337) = -1.78; convert to normal probability = .0375

13.

  1. Sample of n=1000 is less than 5% of population, and np(1-p) is greater than 10 (227.5), so the sampling distribution is normal.
  2. z = .39-.35/(.0151) = 2.64; convert to normal probability = .9959; probability = 0.0041
  3. z = .32-.35/(.0151) = -1.99; convert to normal probability = .0233

14.

  1. Sample of n=1460 is less than 5% of population, and np(1-p) is greater than 10 (355.7), so the sampling distribution is normal.
  2. z = .45-.42/(.0129) = 2.33; convert to normal probability = .9901; probability = 0.0099
  3. z = .4-.42/(.0129) = -1.55; convert to normal probability = .0606

15.

  1. Given an estimate of 300 million Americans, a random sample of 200 fits the first parameter that it be less than 5% of the population. The second parameter, that np(1-p) is greater than 10, is also met - it is 49.82. Therefore, the sampling distribution is normal given N of 300 million and p of 0.47.
  2. z = .5-.47/(.0353) = .85; convert to normal probability = .8023; probability = 0.1977
  3. z = .4-.47/(.0353) = -1.98; convert to normal probability = .0239 - Yes, this would be unusua1 as it would occur only about 2 out every 100 times.

16.

  1. Given an estimate of 300 million Americans, a random sample of 100 fits the first parameter that it be less than 5% of the population. The second parameter, that np(1-p) is greater than 10, is also met - it is 14.76. Therefore, the sampling distribution is normal given N of 300 million and p of 0.82.
  2. z = .85-.82/(.0384) = .78; convert to normal probability = .7823; probability = 0.2177
  3. z = .75-.82/(.0384) = -1.82; convert to normal probability = .0344 - Yes, this would be unusua1 as it would occur only about 3.5 out every 100 times.

17.

  1. Given an estimate of 300 million Americans, a random sample of 500 fits the first parameter that it be less than 5% of the population. The second parameter, that np(1-p) is greater than 10, is also met - it is 118.95. Therefore, the sampling distribution is normal given N of 300 million and p of 0.39.
  2. z = .38-.39/(.0218) = -.46; convert to normal probability = .3228
  3. z = .45-.39/(.0218) = 2.75; convert to normal probability = .9970; z = .4-.39/(.0218) = .46; convert to normal probability = .6772; .9970-.6772 = .3198
  4. z = .42-.39/(.0218) = 1.38; convert to normal probability = .9162; probability = .0838; Yes, this would be unusual as it would occur only about 8 out of every 100 times. *I see the book says this is not unusual - how do you know?