Here is the syntax you can use to check your answers for Section 8.1. (Forward and Backward)
Say you want to know the \(Pr(\bar{X} < 5)\) and \(\bar{X} \sim \mathcal{N}(6,1.5)\)
pnorm(5, mean = 6, sd = 1.5 )
## [1] 0.2524925
Here is the syntax you can use to check if a “Backward” calculation is correct.
Say you know the probability to the left of \(\bar{x}\) = .04 and you want to know what the appropriate \(\bar{x}\) is. You also know that \(\bar{X} \sim \mathcal{N}(6,1.5)\)
qnorm(.04, mean = 6, sd = 1.5)
## [1] 3.373971
15.
- Distribution is normal. Mean is 80, SD is 2
- 0.0668 (c)0.0179
- 0.7969
17.
- Normal distribution. Mean is 64, SD is 4.907
- 0.7486
- 0.4052
19.
- 0.3520
- Normal distrbution, mean is 266 and SD is 3.578
- 0.0465 (d)0.004
- The sample came from a population with a lower average gestation period.
- 93.9
21.
- 0.3085
- 0.0418
- 0.0071
- The increased sample size will decrease the standard deviation resulting in a decreased probability.
- The probability of a random sample is 0.1056. The program isn’t very effective.
- 93.9
23.
- 0.5675
- 0.7291
- 0.8051
- 0.8351
- The likelihood will increase.
Here is the syntax you can use to check your answers for Section 8.2. (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.
- normal distribution, mean is 0l8, SD is 0.046
- 0.1922
- 0.0047
12.
- Normal distribution, mean is 0.65 and SD 0.034
- 0.1867
- 0.0375
13.
- Normal distribution, mean 0.035, Sd is 0.015
- 0.0040
- 0.0233
14.
- Normal distribution, mean is 0.042 and SD is 0.013
- 0.0102
- 0.0606
15.
- Normal distribution, where mean 0.47 and SD is 0.035.
- 0.1977
0.0239 16.
- Normal distribtuion, mean is 0.82 and SD is 0.038
- 0.2177
Probability is 0.0344. So, yes.
17.
(a)Normal distribution where mean is 0.039 and SD is 0.022 (b) 0.3228 (c) 0.3198 (d) 0.0838