7.1

31.

shadenorm(mu = 62, sig = 18, below = 44, col = "blue", dens = 200)

Interpretation 1. 15.87% of the cell phone plans in the U.S are less than $44/mo. and the 15.87% of cell phone plans in the U.S are greater than $80/mo.

Interpretation 2. Type answer here. The probability that a randomly slected cell phone plan with a cost less than $44/mo. is 0.1578. Likewise, the probability that a randomly selected cell phone plan with a cost greater than $80/mo. is 0.1578.

32.

shadenorm(mu = 14, sig = 2.5, above = 17, col = "blue", dens = 200)

Interpretation 1. 11.51% of refrigeratros last for more than 17 years.

Interpretation 2. The probability that a randomly selected refrigerator is more than 17 years old is 0.1151.

33.

shadenorm(mu = 3400, sig = 505, above = 4410, col = "blue", dens=200)

Interpretation 1. 2.28% of full term babies have a birth weight of least 4410 grams.

Interpretation 2. The proboablity that a randomly selected full term new born weighs at least 4410 grams is 0.00228.

34.

shadenorm(mu = 55.9, sig = 5.7, below = 46.5, col = "blue", dens=200)

Interpretation 1. 4.96% of 10 year old males are less than 46.5 inches

Interpretation 2. If a radomly selected 10 year old male is chosen, the probability that he is less than 46.5 inches is 0.0496

35.

Interpretation 1. 19.08% of of all human prenancies will last longer than 280 days.

Interpretation 2. If you randomly select a pregnant woman, the probability that her pregnancy will last longer than 280 days is days.

Interpretation 1 34.16% of pregnancies last between 230 and 260 days

Interpretation 2. If you randomly select a pregnant woman, the probablity that her preganancy will last between 230 and 260 days is 0.3416.

36.

Interpretation 1. In Elenas 40 trials of her experiment, 33.09% of her trials yeild at least 26 miles for one gallon. (or, 33.09 % of trials for miles per gallon are 26 miles per gallon.)

Interpretation 2. If you randomly select one of her trials, there is a .3309 probability that the gallon of that trial will yeild at least 26 miles.

Interpretation 1. 11.07% of trials for miles per gallon are 18-21 miles per gallon.

Interpretation 2. If you randomly slect one of her trials, there is a .1107 probability that the gallon of that trial will yeild 18-21 gallons.

7.2

5.

  1. The area to the left of z= -2.45 is 0.0071.
  2. The area to the left of z= -0.43 is 0.3336
  3. The ara to the left of 1.35 is 0.9115
  4. The area to the left of 3.49 is 0.9998

7.

  1. The area to the right of z= -3.01 is 1-0.0013 = 0.9987
  2. The area to the right of z= -1.59 is 1-0.0559= 0.9441
  3. The area to the right of z= 1.78 is 1-0.9625= 0.0375
  4. THe area to the right of z = 3.11 is 1-0.9991= 0.0009

9.

  1. The area under z= -2.04 and z= 2.04 is 0.9793-0.0207 = 0.9586
  2. The area under z= -0.55 and z= 0 is 0.500-0.2912 = 0.2088
  3. The area under z= -1.04 and z= 2.76 is 0.9971-0.1492=0.8479

11.

  1. (0.0228)+(1-0.9772)= 0.0456
  2. (0.0594)+(1-0.9948)=0.0646
  3. (0.4052)+ (1-0.8849)= 0.5023

13. z score -1.28 is closest to the area of the normal curve to the left being 0.1. (closest is 0.1003)

15. Looking for area to the left to be 0.75 because area to the right is 0.25 (1-0.25=0.75) and that is z score of is 0.67. This is the z score that is closest to the area to the right of the curve being 0.25 (to the right 0.75) and that is 0.7486 more specifically as the area to the left.

17. area of 0.005 has z score -2.57 and -2.58 closest to that area so average of that is -2.575 for z score 1. area of 0.995 has z score 2.57 and 2.58 closest to that area with average of 2.575 so z score 2 is 2.575. The z score for the middle 99% are z score for top and bottom 0.5% because 100%-99%= 1% and top and bottom have to be equal since 99% is in the middle and they have to complete the 100%. 0.005+0.995=1. The area to the right of z score 2 is also 0.005.

33. 9th percentile is 0.09% so need z score with area closest to 0.09 which is -1.34 z score with area 0.0901. This is (-1.34)(7 as standard deviation)+ 50 as mean = 40.62. Thus, the 9th percentile for X is x=40.62.

35. The same process as problem 33, the 81st percentile for X is x = 56.16

37.

shadenorm(mu = 21, sig = 1.0, below = -1000, col = "blue", dens=200)

  1. z = 20-21/1 = -1. Area is 0.1587. less than 20 days so left. So, P(X<20)= 0.1587.
  2. z = 22-21/1 = 1. Area is 0.8413. Greater than 22 days so right. 1-o.8413 = 0.1587. So P(X>22)= 0.1587.
  3. P(19_< X 21_<)= 0.4772. Found area for z score -2 and 0 and found difference.
  4. It would be unusual becasue 18 days gives z score of -3 which is area to the left of 18 days so area is 0.0013 which is a very small probaiblity because that means that only 1 egg out of 1000 hatches in under 18 days.

39.

  1. difference between areas of z scores is 0.8658. So P(1000_<X_<1400) = 0.8658.
  2. P(X<1000)= 0.0132
  3. P(X>1200) = 0.7019
  4. P(X<0.1125) = 0.1230.
  5. Find z score, then area, and percetile is the percent version of the area. Here it is 96th percentile. The z score was 1.81 and the P(X<1475) = 0.9649.
  6. Percentile is 4th percentile and P(X<1050)=0.0359.

41.

  1. P(X>270) = 0.4013. Found score of 0.25, subtract that from 1 and then that area is the probability of more than 270 days.
  2. P(X<250)= 0.1587
  3. P(240_<X_<286)= 0.7590.
  4. P(X>280) = 0.1894
  5. P(X_<245)= 0.0951
  6. P(X<224) = 0.0043 and that is about 4 births in 1000 so yes a very preterm baby is unusual.

43.

  1. P (X<24.9) = 0.0764
  2. P (X<24.85 0r X>25.15) = 0.0324
  3. 0.0324 is 3.24% and 3.24% of 5000 is 162 that he should expect to discard.
  4. You need to make 11.504 because P(24.9_<X_<25.1)= 0.8472 so only 84.72% will be make correctly. So, you need 11,504 rods made so that 10,000 rods can be made correclty because given the probability, 84.72% of 11,504 will yeild you the desired 10,000 correct rods.