7.4 - 8.1 Test Notes
Binomial Probability formula
(nCx)(P^x)(^n-x) ex. n = 50, p = 0.3, x = 20 Calculator: (50 nCx 20) (0.3^20)(0.7^37) (note: nCx is [MATH] [nCr])
A simple random sample of size n= 56 is obtained from a population with μ= 78 and σ= 77. Does the population need to be normally distributed for the sampling distribution of x-bar to be approximately normally distributed? Why? What is the sampling distribution of x-bar?
No because the Central Limit Theorem states that 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 or approximately normal with µx = 78 and σx = 0.935. σx = 7/√56 = 0.935.
What is the area under the normal curve that describes the probability that more than 52 households have a gas stove?
The area to the right of 52.5
The distribution of the sample mean, x-bar, will be normally distributed if the sample is obtained from a population that is normally distributed, regardless of the sample size.
True.
The standard deviation of the sampling distribution of xbar, denoted σx, is called the __a__ __b__ of the __c__.
a. Standard b. error c. mean
In a binomial experiment with n trials and probability of success p, if__a__, __b__,__c___ the binomial random variable X is approximately normal with u(x) = __d__ and q(x) = __e__
a. np (1 - p) b. ≥ c. 10 d. np e. √np(1 - p)
According to the American Red Cross, 7% of people in the United States have blood type O - negative. What is the probability that, in a simple random sample of 500 people in the U.S., few than 30 have blood type O-negative?
step 1: verify np(1 -p) ≥ 10. 500 (0.07) (1 - 0.07) = 32.55 ≥ 10. P(X < 30) = P(X ≤ 29) mean = 500(0.07) = 35 Std = √500(0.07) (1-0.07) = √32.55 ≈ 5.71 Step 2: [2nd] [Distr] [normalcdf] lower: -1E99 upper: 29.5 mean: 35 Std: 5.71 first value: 0.1677 check: [2nd] [Distr] [binomcdf] trials: 500 p:.07 x value: 29 second value: 0.1677 answer: P(X < 30) = 0.1677
Determine μx and σx from the given parameters of the population and sample size. μ= 71, σ= 99, n= 81
ux = 71 σx = 9/√81 = 1.
To cut the standard error of the mean in half, the sample size must be doubled.
False. The sample size must be increased by a factor of four to cut the standard error in half.
To compute probabilities regarding the sample mean using the normal model, what size sample would be required?
The sample size needs to be greater than or equal to 30.
Suppose that X is a binomial random variable. To approximate P (3 great or equal X great or equal 7) using the normal probability distribution, we compute P(3.5 great or equal X great or equal 7.5)
The statement is false.
According to the Gallup Organization, 65% of adult Americans are in favor of the death penalty for individuals convicted for murder. Erica selects a random sample of 1000 adult Americans in Will County, Illinois, and finds that 630 of them are in favor of the death penalty for individuals convicted of murder. a. Assuming that 65% of adult Americans in will county are in favor of the death penalty, what is the probability of obtaining a random sample of no more than 630 adult Americans in favor of the death penalty from a sample of size 1000? b. Does the result from part (a) contradict the Gallup Organizations findings?
a. step 1: verify np(1 -p) ≥ 10. 1000(.65)(.35) = 227.5 ≥ 10. P (X ≤ 630) ≈ area to the left of x = 630.5 Step 2: Mean = np = 1000(.65) = 650. σx = √1000(0.65)(0.35) ≈15.083 Step 3: [2nd] [Distr] [normalcdf] lower: -1E99 upper: 630.5 mean: 650 Std: 15.083 estimate probability: 0.098 check: [2nd] [Distr] [binombdf] trials: 500 p:0.65 x value: 630 second value: 0.0984 answer: If we obtained 100 different samples of size 1000, we would expect about 10 (10%) to result in 630 or fewer Americans favoring the death penalty if the true proportion is 0.65. Because the results obtained are not unusual (less than 0.05) under the assumption that p = 0.65, it does not contradict.
The reading speed of second grade students in a large city is approximately normal, with a mean of 89 words per minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f). a. What is the probability a randomly selected student in the city will read more than 95 words per minute? b. Interpret this probability. Select the correct choice below and fill in the answer box within your choice.
a. 95 - 89/10 = 0.6. The area to the right of z = 0.6 is 1 - 0.7257 = 0.2743. (note: 0.7257 was found on z-chart under 0.6). b. If 100 different students were chosen from this population, we would expect 27 to read more than 95 words per minute. (note: 27 was given through 100 - 73).
A simple random sample of size n=81 is obtained from a population with μ=85 and σ=9. a. Choose the correct description of the shape of the sampling distribution of x-bar. b. Find the mean and standard deviation of the sampling distribution of x-bar. c. (x-bar > 86.5)
a. The distribution is approximately normal. b. ux = 85, σx = 9/√91= 1. c. [2nd] [distr] [normalcdf] lower: 86.5 upper: 1E99 u: 85 σ: 1 value: 0.0668
Suppose a simple random sample of size n is drawn from a large population with mean μ and standard deviation σ. The sample distribution of over bar x has mean μ=__a___ and standard deviation σ=__b___.
a. µ b. σ/√n
The mean weight gain during pregnancy is 30 lbs, with a std. dev. of 12.9 lbs. Weight gain during pregnancy is skewed right. an obstetrician obtains a random sample of 35 low-income patients and determines their mean weight gain during pregnancy was 36.2 lbs. Does this result suggest anything unusual?
ux = 30 σx = 12.9/√35 = 2.180 P(µ≥36.2) Calculator: [2nd] [distr] [normalcdf] lower: 36.2 upper: 1E99 u: 30 σ: 2.180 answer value: P(µ≥36.2) = 0.0022
he data in the table represent the ages of the winners of an award for the past five years. Use the data to answer questions (a) through (e). {41, 41, 51, 65, 63} a. Compute the probability that the sample mean is within 4 years of the population mean age. The probability is
µ = 261/5 = 52.2 -Probability: order the sample means and how many times it occurs. -mean of the sampling distribution is 52.2. a.
The IQ, X, of humans is approximately normally distributed with mean u = 100 and standard dev. q = 15. Compute the probability that a simple random sample of size n = 10 results in a sample mean greater than 110. That is, Compute P( mean > 110)
µx = 100 σx = 15/√10 = 4.743 P(µ > 110) Calculator: [2nd] [distr] [normalcdf] lower: 110 upper: 1E99 u: 100 σ: 4.743 answer value: P(µ>100) = 0.0175
