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

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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?

Answer

First we standardize the score with
z=(x-u)/o
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:

z=(x-100)/15
15z+100=x

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

Answer: 131

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
√(576)=24

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%

Answer: 57.62%

Empirical Rule: 68-95-99.7
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

Answer: 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?

Equation: √(1/Sample)(σ)
Mean: 18

16-18=2^2= 4
19-18=1^2= 1
18-18=0^2=0
17-18=-1^2=1
20-18=2^2=4
18-18=0^2=0

Then add them all up 4+1+1+4= 10

√(1/5(10) = 1.414

Answer: 1.414

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