7.2
- type answer here. Standard normal distribution
- type answer here. Z(alpha)
- type answer here. 0.3085
- type answer here. 0.4207
- type answers here. 0.0071
- type answers here. 0.3336
- type answers here. 0.9115
- type answers here. 0.9998
- type answers here. 0.9987
- type answers here. 0.9441
- type answers here. 0.0375
- type answers here. 0.0009
- type answer here. z=0.67
- type answer here. z1=-2.575 z2=2.575
- type answer here. z1=-1.88 z2=1.88
- type answer here. 0.9838
shadenorm(mu=50, sig=7, above = 35,col = "blue",dens = 150)

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

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

- type answers here. 0.8658
- type answers here. 0.0132
- type answers here. 0.7019 of the bags have more than 1200 chocolate chips
- type answers here. 0.1230 of the bags have fewer than 1125 chocolate chips
- type answers here. A bag that contains 1475 chocolate chips is at the 96th percentile
- type answers here. A bag that contains 1050 chocolate chips is at the 4th percentile.
8.1
- type answer here. sample distribution
- type answers here. mean, standard deviation divided by the square root of n
- type answer here. standard error of the mean
- type answers here. true
- type answers here. false
- type answers here. false
- type answers here.(Need the shape, center and spread) ̅ The sampling distribution of x-bar is normal with mean =30 and standard deviation = 8 devided by the square root of 10 is equal to about 2.530.
- type answers here. No the population does not need to be normally distributed for the sampling ditribution of x-bar to be approximately normally distributed because regardless of the shape of the underlying population, the sampling distribution of x-bar becomes approximately normal as the sample size, n, increases. The sampling distribution of x-bar is normal with a mean of = 50 and standard deviation = 8 divided by the square root of 40 is equal to about 1.265.
- type answers here. mean = 80, standard deviation = 2
- type answers here. mean x-bar= 64, standard deviation x-bar= 18/square root of 36 which is equal to 3.
- type answers here.(Need the shape, center and spread) mean is approximately normal with mean x-bar = 80 and standard deviation x-bar = 20
- type answers here. 0.0668. If we take 100 simple random samples of size n = 49 from a population with mean = 80 and standard deviation = 14, then about 7 of the samples will result in a mean that is greater than 83.
- type answers here. 0.0179. If we take 100 simple random samples of size n = 49 from a population with a mean = 80 and standard deviation = 14, then about 2 of the samples will result in a mean that is less than or equal to 75.8.
- type answers here. =0.7969. If we take 100 simple random samples of size n = 49 from a population with mean = 80 and standard deviation = 14, then about 80 of the samples will result in a mean that is between 78.3 and 85.1.
- type answers here. =0.3520. If we randomly select 100 human pregnancies, then about 35 of the pregnancies will last less than 260 days.
- type answers here. The sampling distribution of x-bar is normal with mean (center) = 266 and standard deviation (spread) = 16/square root of 20 equally to about 3.578.
- type answers here. =0.0465. 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.
- type answers here. =0.0040. 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.
- type answers here. This result would be unusual. The sample most likely came from a population that had a mean gestation period that is less than 266 days.
- type answers here. =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.