HW6_skeleton

7.2

• 1. Standard Normal Distribution
• 2. 1-Zα
• 3. .3085
• 4. 0
• 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 area to the left of z = 1.35 is 0.9115

  4. The area to the left of z = -3.49 is 0.9998

• 1. The area to the right of z = -3.01 is 1-0.0013 = 0.9937

  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

• 15. The z-score is 0.67. (Area to the left of unknown z-score =’s 0.75. The area in the interior of the standard normaltable that’s closest to 0.7500 is 0.7486.)
• 17. The scores (after averaging) are -2.575 and 2.575
• 18. 1. P(Z ≤ -Z sub 0) =1-0.94/2 = 0.03
  2. P(Z ≥ Z sub 0) = 1-0.94/2 = 0.03

  Answer: from (1) and (2) the Z-scores are -1.88 and 1.88 approximately.

• 23. The area to the left is 0.0162, so P(X > 35) = 1 - 0.0162 = Answer: 0.9838

shadenorm(mu=50, sig=7, above = 35,col = "blue",dens = 150)

• 24. P(Z > 2.14) represents the area to the right of Z = 2.14. 1 - (area to the left of Z = 2.14) = 1 - .9838 = 0.0162. Thus, P(X > 65) = P(Z > 2.14) Answer = 0.0162

shadenorm(mu=50, sig=7, above = 65,col = "blue",dens = 150)

• 25. Z = 45-50/7 = (approximately) -0.71. The area to the left is 0.2389, so P(X ≤ 45) = 0.2389

shadenorm(mu=50, sig=7, below = 45,col = "blue",dens = 150)

• 1. P(-2.22 ≤ Z ≤ 1.17) = P(Z ≤ 1.17) - P(Z≤-2.22) =0.8790 - 0.0132 Thus the probability that a randomly selected 18-ounce bag of Chips Ahoy! Cookies contain between 1,000 and 1,400 choc. chips is 0.8650

  2. The area to the left of z = -2.22, which is 0.0132, meaning P(X < 1,000) = 0.0132. Thus the probability that a randomly selected 18-ounce bag of Chips Ahoy! Cookies contain fewer than 1,000 choc. chips is 0.0132

  3. Area to the left of Z = -0.53 is 0.2981, so P(X > 1,200) = 1 - 0.2982 = 0.7019. So, the proportion of 18-ounce bags of Chips Ahoy! cookies that contains more than 1,200 choc. chips is 0.7019 or 70.19%

  4. Area to the left of Z = -1.16 is 0.1230, so P(X > 1,125) = 0.1230. So, the proportion of 18-ounce bags of Chips Ahoy! cookies that contains fewer than 1,125 choc. chips is 0.1230 or 12.30%.

  5. The area to the left of z = 1.81 is 0.9649, so P(X < 1475) = 0.9649. So, An 18-ounce bag of Chips Ahoy! Cookies that contains 1,475 choc. chips is at the 96th percentile.

  6. The area to the left of z = -1.80 is 0.0359, so P(X < 1,050) = 0.0359, so P(X < 1,050) = 0.0359. So, an 18-ounce bag of Chips Ahoy! Cookies that contains 1,050 choc chips is at the 4th percentile.

8.1

• 1. Sampling Distribution
• 2. mu, standard deviation devided by square root of n
• 3. The Standard Error of the Mean
• 4. True: If the population is normal, then the distribution of the sample mean is also normal.
• 5. False: if the sample is drawn from a population that is not normal, then the distribution of the sample mean will not necessarily be normally distributed
• 6. False: the sampling distribution of mean becomes approximately normal as the sample size, n, increases. Therefore, in order to cut the standard error of the mean in half, the sample size is quadrupled.
• 7. The sampling distribution is normal. The mean of the sample distribution is 30. The standard deviation is 8/Square Root of 10.
• 8. No it does not due to the central limit theorom which describes that regardless of the shape of the underlying population and the sampling distribution of x-bar it will become approximately normal as the sample size (n) increases. The standard deviation of the sampling distribution of sample mean is: 0.632
• 9. 14/7=2
• 10. μx-bar = 64 σx̅ = 3
• 1. μx-bar = μ = 80; σx̅=14/7=2. We are not specificially told that the population is normally distributed, but we do have a large sample size (n = 49≥30). Therefore, by using the Central Limit Theorem, we can say that the sampling distribution of x-bar is approximately normal.

  2. 1-0.9332=0.0668. If we were to take 100 simple randonm ramsples of size n=49 from a population with μ=80 and σ=14, then about 7 of the samples wull result in a mean that is greater than 83.

  3. P(x-bar ≤ -2.10) = 0.0179. If we were to take 100 simple random samples of size n=49 with μ=80 and σ=14, then about 2 of the amples will result in a mean that is lesss than or equal to 75.8.

  4. P(-0.85 < Z < 2.55) =0.9946-0.1977=0.7969. If we were to take 100 simple random samples of n=49 with μ=80 and σ=14, then about 78 of the samples will result in a mean that is between 78.3 and 85.1.

• 1. 1-P(Z<0.24) =1-0.5948 =0.4052 So if we take 100 simple random samples of n=100 human pregnancies, then about 35 of the pregnancies would last less than 260 days.

  2. Since the length of human pregnancies is normally distributed, the sampling distribution of x-bar is normal with μ x-bar=266 and σx-bar≈3.578

  3. P(Z ≤ -1.68) =0.0465 So if we take 100 simple random samples of size n =20 human pregnancies, then about 5 of the samples will result in a mean gestation period of 260 days or less.

  4. P(Z ≤ -2.65) = 0.0040 SO if we take 1000 simple random samples of size n=50 human pregnancies, then about 4 of the samples will result in a mean gestation period of 260 days or less.

  5. The answers are going to differ. Part (d) indicates that the result would be an unsual observation. Therefore, we can conclude that the sample likely came from a population whose mean gestation period is less than 266 days.

  6. P(-2.42 ≤ Z ≤ 2.42) =0.9922-0.0078 =0.9844 So if we take 100 simple andom samples of size n = 15 human pregnancies, then about 98 of the samples will resultt in a mean gestation period between 256 days and 276 days (inclusive).