HW6_skeleton

7.2

• 1. Standard normal distribution
• 2. ?? or alpha
• 3. 0.3085
• 4. 0.4207
• 1. 0.0071

  2. 0.3336

  3. 0.9115

  4. 0.9998

• 1. 1 - 0.0013 = 0.9987

  2. 1 - 0.0559 = 0.9441

  3. 1 - 0.9625 = 0.0375

  4. 1 - 0.9991 = 0.0009

• 15. Area left of the unknown z-score is 1 - 0.25 = 0.75, closest value to this is 0.7486 which corresponds to z = 0.67
• 17. Area left of z1 is 0.005, area left of z2 is 0.995. The thing closest to 0.005 is 0.0049 and 0.0051. Use average of both scores to give us z1 = -2.575. The area in the interior of the table are closest to 0.995 are 0.9949 and 0.9951. Average of these gives us z2 = 2.575.
• 18. Area left of z1 is 0.030, area left of z2 is 0.97. Therefore z1 = 0.5120. On the other hand, z2 = 0.8340
• 23. P(X>35) = 0.9838

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

• 24. P(X>65) = 0.0166

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

• 25. P(X???45) = 0.2389

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

• 1. P(1000???x??? 1400)=0.8790-0.0132 = 0.8658

  2. z=-2.22 is 0.0132. Therefore P(x<1000) = 0.0132

  3. Z=-0.35 is 0.2981. So P(x>1200)=1-0.2981=0.7019. Therefore proportion of 18 ounce bags of cookies that contain more than 1200 chocolate chips is 0.7019 or 70.19%.

  4. Z=-1.16 is 0.1230. So P(x<1125)=0.1230. Therefore the proportion of 18 ounce bags of cookies that contains fewer than 1125 choclate chips is 0.1230 or 12.30%

  5. z=1.81 is 0.9649 so P(x<1475) = 0.9649. Therefore an 18 ounce bag of cookies that contains 1475 chocolate chips is at the 96th percentile.

  6. z=-1.8 is 0.0359 so P(x<1050)=0.0359. An 18 ounce bag of cookies that contains 1050 chocolate chips is at the 4th percentile.

8.1

• 1. Sample distribution
• 2. ??, ??/???n
• 3. Standard error, mean
• 4. True
• 5. False, if the sample is drawn from a population that isn’t normal then the distribution of the sample mean will not necessarily be normally distributed.
• 6. False.
• 7. The sampling distribution is normal. The mean of the sample distribution x-bar is 30. The standard deviation is 8 divided by the square root of 10.
• 8. type answers here.
• 9. Mean = 80, standard deviation = 2.
• 10. Mean = 64, standard deviation = 3.
• 1. Mean is 80 and standard deviation is 14. Therefore the mean of of the sampling distribution of x-bar is 80 and the standard deviation of x-bar is 2. We are not given information on whether the population is normally distibuted but we have a large sample size. Therefore we can use the central limit theorem to say that the sampling distribution of x-bar is approximately normal.

  2. 1-P(z???1.50) = 1-0.9332=0.0668. So if we take 100 simple random samples of size n=49 from a population with mean of 80 and a standard deviation of 14 then about 7 of these samples will result in a mean that is greater than 83.

  3. P(z???(75.8-80/2)) = P(Z???-2.1)= 0.0179. If we take 100 simple random samples of size n=49 from a population with mean of 80 and standard deviation of 14 then about 2 of the samples wll result in a mean less than or equal to 75.8.

  4. P(((78.3-80)/2) < z < ((85.1-80)/2)) = P(-0.85<z<2.55) = 0.9946-0.1977 = 0.7969. If we take 100 simple random samples of size n=49 from a population with a mean of 80 and standard deviation of 14 then about 78 of the samples will reslt in a mean that is between 78.3 and 85.1.

• 1. P(z<((260-266)/16) = P(Z<-0.38) = 0.3520. Out of 100 pregnancies 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?? is normal with the mean = 266. and standard deviation equal to 3.578.

  3. P(z???-1.68) = 0.0465. If we take 100 simple random samples of n=20 human pregnancies then 5 of the samples will result in a mean gestation period of 260 days or less.

  4. P(z???-2.65) = 0.0040 If we take 1000 simple random samples of size n=50 human pregnancies then 4 of the samples will result in a mean gestation of 260 days or less.

  5. The answers here will generally vary. The previous part shows that the result would be an unusual observation. Therefore we can conclude that the sample likely came from a population whos mean gestation period is less than 266 days.

  6. P(02.42 ??? Z ??? 2.42) = 0.9922 - 0.0078 = 0.9844. If we take 100 simple random samples of size n = 15 human pregnancies, then about 98 of the samples will result in a mean gestation period between 256 and 276 days, inclusive.