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Assuming that a score of 135 was earned on a test with a standard deviation of 15 (not 16 or 24) (e.g. Wechsler, as opposed to Stanford-Binet or Cattell), said score places one at the 99th percentile (assuming a mean score of 100; globally, mean IQ is approximately 88, circa 2015, placing a score of 135 above the 99.9th percentile [one per one-thousand people]). There are some intellectual (high-IQ) societies that accept only individuals who have scored at or above the top 1% (mean 100); the largest and oldest of these is Intertel.
Assuming that you mean additive opposite, the answer is +11.Assuming that you mean additive opposite, the answer is +11.Assuming that you mean additive opposite, the answer is +11.Assuming that you mean additive opposite, the answer is +11.
I am assuming you had your IQ test professionally as opposed to someone's personal test on the internet. I am 19 years old and have a 105 IQ. IQ and IQ tests can vary with age so that factor is important. Most people my age score between 90 and 109 IQ on their test so 105 is a good average score if your about my age. You are most likely at the 63rd percentile (I am). Percentile is NOT the same as percentage. Being at the 63rd percentile means out of 100 other people who took that test, you rank 63rd. You are NOT smarter than 63% of the population as the internet would have you believe. Internet IQ tests are highly inaccurate, I know because I've tried both internet IQ tests and had a professional IQ test done with the results of the professional greatly differing from the internet. You tester should have gone over any questions you had when you got your results. Provided you did the job properly and had you test professionally done. Which you should do, just incase you missed it.
Assuming you mean 10/20, (10/20)/(4-5)=0.5/-1=-1/2 Assuming you mean 10.20, 10.2/-1=-10.2 Assuming you mean 10-20, (10-20)/(-1)=(-10)/-1=10
well assuming you want the board feet and assuming it is a square you would multiply 8 by 5 and get - 40 square board feet
Assuming a normal distribution of incomes: 2672z = ( 2672 - 3036 ) / 950 = -0.383157895Pr{z
Assuming a normal distribution 68 % of the data samples will be with 1 standard deviation of the mean.
The answer will depend on the distribution of the weights. There is no basis for assuming that the distribution is normal, or even symmetrical.
(I am answering the question assuming it can fairly be rephrased "How does an IQ of 106 rate/rank on an IQ scale") By definition, an IQ of 100 is "average". Therefore a score of 106 is slightly above average. More specifically, using the WAIS/Stanford Binet definition wherein a standard deviation of 0.25 corresponds to 4 IQ points, an IQ of 106 roughly corresponds to a standard deviation of 0.38/0.39. What all that means is that an IQ of 106 roughly corresponds to a percentile rank within the population of about 65%, or, in other words, a person with an IQ of 106 is "smarter" than 65% of the population. By one accounting, the average college graduate has an IQ of 116 (and a percentile rank of 84%).
Assuming a normal distribution, the proportion falling between the mean (of 8) and 7 with standard deviation 2 is: z = (7 - 8) / 2 = -0.5 → 0.1915 (from normal distribution tables) → less than 7 is 0.5 - 0.1915 = 0.3085 = 0.3085 x 100 % = 30.85 % (Note: the 0.5 in the second sum is because half (0.5) of a normal distribution is less than the mean, not because 7 is half a standard deviation away from the mean, and the tables give the proportion of the normal distribution between the mean and the number of standard deviations from the mean.)
Assuming var is variance, simply square the standard deviation and the result is the variance.
The answer will depend on the underlying distribution, which is not specified. The normal distribution, although often used at school is not really appropriate because it allocates a positive probability to the event that the lifespan is negative. That is clearly an impossible event! A more appropriate distribution is often the lognormal.Anyway, assuming that you are of school age, and therefore assuming a normal distribution,P(Lifespan < 136) = 0.37P(Lifespan > 150) = 0.20
n probability theory and statistics, thestandard deviation of a statistical population, a data set, or a probability distribution is the square root of itsvariance. Standard deviation is a widely used measure of the variability ordispersion, being algebraically more tractable though practically less robustthan the expected deviation or average absolute deviation.It shows how much variation there is from the "average" (mean). A low standard deviation indicates that the data points tend to be very close to the mean, whereas high standard deviation indicates that the data are spread out over a large range of values.For example, the average height for adult men in the United States is about 70 inches (178 cm), with a standard deviation of around 3 in (8 cm). This means that most men (about 68 percent, assuming a normal distribution) have a height within 3 in (8 cm) of the mean (67-73 in (170-185 cm)) - one standard deviation, whereas almost all men (about 95%) have a height within 6 in (15 cm) of the mean (64-76 in (163-193 cm)) - 2 standard deviations. If the standard deviation were zero, then all men would be exactly 70 in (178 cm) high. If the standard deviation were 20 in (51 cm), then men would have much more variable heights, with a typical range of about 50 to 90 in (127 to 229 cm). Three standard deviations account for 99.7% of the sample population being studied, assuming the distribution is normal (bell-shaped).
Assuming that a score of 135 was earned on a test with a standard deviation of 15 (not 16 or 24) (e.g. Wechsler, as opposed to Stanford-Binet or Cattell), said score places one at the 99th percentile (assuming a mean score of 100; globally, mean IQ is approximately 88, circa 2015, placing a score of 135 above the 99.9th percentile [one per one-thousand people]). There are some intellectual (high-IQ) societies that accept only individuals who have scored at or above the top 1% (mean 100); the largest and oldest of these is Intertel.
It allows you to understand, or comprehend the average fluctuation to the average. example: the average height for adult men in the United States is about 70", with a standard deviation of around 3". This means that most men (about 68%, assuming a normal distribution) have a height within 3" of the mean (67"- 73"), one standard deviation, and almost all men (about 95%) have a height within 6" of the mean (64"-76"), two standard deviations. In summation standard deviation allows us to see the 'average' as a whole.
BMI varies from person to person and one measure of this variability is the standard deviation. Assuming the BMI is approximately normally distributed, only around 0.135% of the people will have results that are -3 sd or lower.
Assuming that we have a Normal Distribution of Data, approx. 65% of the data will fall within One Sigma.