We thoroughly check each answer to a question to provide you with the most correct answers. Found a mistake? Let us know about it through the REPORT button at the bottom of the page. Ctrl+F (Cmd+F) will help you a lot when searching through such a large set of questions.
A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa?
First we standardize the score with
where z is the distance in standard deviations from the mean, x is the score to be standardized, u is the mean of the population, and o is the standard deviation of the population [or sample].
Now, we can suspect from a percentile rank chart that the upper 2% is 2 standard deviations from the mean. In fact, however there is approximately only 100-2.27=97.73 within two standard deviations above the mean. Chances are, however, that your instructor only wants 2 standard deviations above the mean:
This gives a score, x, as a function of its distance from the mean [z]. So if we want 2 standard deviations above the mean, 15(2)+100=130. You will need an IQ of 130 to get into MENSA accordingly. Strictly speaking, the score would be a little higher than 130 (in order for it to be ahead of 98% of the population) due to the reasoning above.
A person must score in the upper 2% of the population on an IQ test to qualify for membership in Mensa, the international high-IQ society (US Airways Attache, September 2000). There are 110,000 Mensa members in 100 countries throughout the world (Mensa International website, January 8, 2013).
If IQ scores are normally distributed with a mean of 100 and a standard deviation of 15, what score must a person have to qualify for Mensa (to the nearest whole number)?
|Find z for p (z ≤ ?) = .98; z=2.05|
x = 2.05*15+100=130.75
|After computing a confidence interval, the user believes the results are meaningless because the width of the interval is too large. Which one of the following is the best recommendation?||Increase the sample size.|
|The average age in a sample of 190 students at City College is 22. As a result of this sample, it can be concluded that the average age of al the students at City College…||could be larger, smaller, or equal to 22|
|The measure of location which is the most likely to be influenced by extreme values in the data set is the..||Mean(µ)|
|The variance of a sample of 169 observations equals 576. the standard deviation of the sample equals…||STD(σ) = √(variance)|
Variance = STD^2
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
What percent of players weigh between 180 and 220 pounds?
|Mean(µ) = 200|
STD(σ) = 25
220-200/25 = 20/25 = .80
Then go find the z-score standard normal table for .80 which is .7881
180-200/25 = -20/25 = -.80
Then go find the z-score standard normal table for -.80 which is .2119
.7881-.2119 = .5762 = 57.62%
1 STD 68 above or below mean
2 STD 95 above or below mean
3 STD 99.7 above or below mean
|Since the Sample size is always smaller than the size of the population , the sample mean…||Can be smaller, larger, or equal to the population mean|
|A simple random sample of 64 observation was taken from a large population. The sample mean and the sample standard deviation were determined to be 320 and 120 respectively. The standard error of the mean is..||STD = 120|
n = 64
x(bar) = 320
120/√(64) = 15
The following information was collected from a simple random sample of a population…. 16 19 18 17 20 18
The point estimate of the population standard deviation is?
Then add them all up 4+1+1+4= 10
√(1/5(10) = 1.414