8.1

## Here is the syntax you can use to check the probabilities you look up are correct.

## Say you want to know the Pr(X < 5) and X is Normal with a mean of 12 and standard deviation 4

pnorm(5, mean = 12, sd = 4 )
## [1] 0.04005916

15.

  1. the distribution is normal
  2. P( Z= 1.5 = 0.9332, SO 1 - 0.9332 = 0.0668 - if we take 100 simple random sample, about 7 will be greater than 83
  3. z= -2.1 = 0.0179 - if we take a simple random sample of 100, about 2 of the sample will be less than or equal to75.8
  4. P(78.3 < z < 85.1) P(-1.85 < Z < 2.55) P( 0.1977 < Z < 0.9946), SO 0.9946 - 0.1977 = 0.7969 If we take 100 simple random sample, about 80 will have a mean between 78.3 and85.1

17.

  1. The population must be normally distributed in order to calculate probabilituies involving the sample mean; if the populatuon is normally distributed, then the sampling distribution will be the same
  2. Z= 0.6725 = 0.7486 - IF you have 100 simple randim samples of size n= 12 then 75 samples will be less than 67.3
  3. Z = 0.2445 = 0.5948, 1 - 0.5948 = 0.4052 - SO if you take a simple random sample of 100 with n= 12 then 41 samples will be equal or greatre than 65.2

19.

(a)Z = -0.375, ROUND IT TO 0.38 = 0.3520; If we take 100 simple randim sample of human pregnancies then 35 of pregnancies will be less than 260 days.

  1. The sample distribution of x is normal with mean = 266 and standart deviation = standard deviation of the population divided by the square root of teh sample size.

  2. Z= -1.677 ROUND IT TO -1.68 = 0.0465, SO If we take 100 simple random sample, then 5 of the sample will have a gestation that would last 260 days or less.

  3. Z = -2.65 = 0.0040, So out of 1000 simple random sample with n = 50 the 4 out of the sample will have a gestation that would last 260 days or less.

  4. The result would be unusual, so the sample likely came from a population whose mean was less than 266 days

  5. P( 256 ??? x ??? 276) P (0.0078 ??? x 0.9922) 0.9922 - 0.0078 = 0.9844, SO out of 100 simple random saple, 98 would result i na mean gestation period between 256 and 276 days.

21.

  1. P(x>95) = Z= 95-90/10 = 0.5 = 0.6915, SO (1-0.6915 = 0.3085). So out of 100 simple random sample, 31 student will read more than 95 words per minute
  2. Z = 1.73 = 0.9582, SO (1-0.9582 = 0.0418), SO OUT OF 100 Simple random sample with n = 12, the around 4 students will read more than 95 words per minute (c)Z = 2.45 = 0.9929 = SO (1- 0.9929 = 0.0071) - Out of 100 simple random sample with n = 24, 7 of the students will read more than 95 words per minute
  3. Increasing the sample size n will decrease the probability because the sample standard deviation decreases as n increases
  4. the readu=ing of 92.8 wpm is not unusual the probability of sample x being more or equal than 92.8 = 0.1056 ,so basically only 10 out of 100 would have done well; that sais, we can conclude that the new program is NOT better than the old one.

23.

  1. Z = 0.1750 = 0.5675 - Simple sample of 100 months, then 57 months will have positive rates return. (b)Z = 0.61 = 0.7291, SO out of 100 simple random sample, with n = 12, then 73 of the sample wil lhave a mean monthly rate of return positive.
  2. Z = 0.86 = 0.8051, SO 100 Simple random sample with n = 24 then 81 sample will have a mean monthly rate of return positive.
  3. Z= 1.05 = 0.8531, So out of 100 simple s=random sample with n = 36, then 85 of the sample will have monthly rate of return positive.
  4. As the number of months increasses (n increases) you’re more likely to earn a positive rate of retun