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. n=75 is than 5 of 10,000 and np(1-p)= 12 which is > than 10; the population distribution of p is approximately normal if mean = 0.8 and O = 0.046
  2. Z = 0.87 = 0.8078 (1- 0.8078 = 0.1922); About 19 random sample out of 100 of size n= 75 will result with 63 or more individual with the same characteristic
  3. Z = -2.608 = 0.0047; About 5 out of 100 random sample of size n=75 will result in 51 or less individual with the characteristic.

12.

  1. Tn= 200 is less than 5% of 25,000; np (1-p) = 45.5 which is > than 10[ the distribution of p is approximately normal if mean = 0.65 and o = 0.033
  2. z= 0.91 = 0.8186 (1-0.8186= 0.1814); About 18 out of 100 random sample of size n= 200 will result with 136 or more ins. with characteristic
  3. z= -2.73 = 0.0032; about 3 out of 100 random sample of size n= 200 will result in 118 or less Individual with the characteristic.

13.

  1. n= 1000 which is lss than 5% of 1,000,000; np(1-p) = 227.5 wich is > than 10; so the ditribution is approximately normal if mean = 0.35 and o = 0.015
  2. p = x/n = 390/1000 = 0.39; z= 2.66 = 0.9961 (1-0.9961 = 0.0039); About 4 out of 100 random sample of n = 100 will result in 390 or more individual with the characteristic.
  3. Tp=x/n = 320/1000 = 0.32; Z= -2 = 0.0228; out 2 of the random sample will result in 320 or less individual with the characteristic

14.

  1. n = 1460 is less than 5% of N=1,500,000; np(1-p) = 355.656; The distribution is approximately normal if mean = 0.42 and o = 0.013
  2. p= x/n = 657/1460 = 0.45; Z=2.31 = 0.9896 (1-0.9896 = 0.0104); 1 out of 100 wil lresult in 657 or more ins. with the characteristic.
  3. Z = -1.54 = 0.0618; About 6 random sample of n = 1460 will result in 584 or less individual with the characteristic.

15.

  1. The distribution is approximately normal if mean = 0.47 and o = 0.035
  2. Z = 0.85 = 0.8023 (1-0.8023 = 0.1977); 20 out of 100 random sample of n =200 wil lresult in more than 100 ins. that can order a meal in a foreigh language.
  3. p =x/n = 80/100 = 0.4; z=-2 =0.0228; 2 out of 100 random sample of n 200 wil lresult in 80 or less who can order a meal i na foreigh language. This is unusual.

16.

  1. The distribution is approximatelly normal if mean=0.82 and o = 0.038
  2. p=x/n = 0.75 Z=0.78 = 0.7823 (1-0.7823 = 0.2177); 21 out of 100 random sample with n = 100 will result with more than 85 individual that are satisfied with their lives.
  3. Z=-1.84 = 0.0329; 3 out of 100 random sample with n=100 wil lresult in 75 or less individual that are satisfied with the lives. Unusual.

17.

  1. The distribution is approximately normal if mean=0.39 and o = 0.022
  2. Z=-0.46 = 0.3228; 32 out of 100 random sample with n = 500 wil lresult with less than 190 ind. that believe marriageolete
  3. z = 0.46 - 0.6772; AND z=2.73 = 0.9968 0.9968 - 0.6772 = 0.3196; 32 out of 100 random sample with n = 500 wil lresult in between 200 or 225 adult that believe marriage is obsolute
  4. p=x/n=0.42 z = 1.37 = 0.9147 (1-0.9147 = 0.0853); 8 out of 100 random sample of n = 500 will result in 210 or more adult who believe marriage is obsolete.