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 Then add them all up 4+1+1+4= 10 √(1/5(10) = 1.414 Answer: 1.414 |