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Why does a researcher want to go from a normal distribution to a standard normal distribution?
Wiki User
∙ 14y ago
A researcher wants to go from a normal distribution to a standard normal distribution because the latter allows him/her to make the correspondence between the area and the probability. Though events in the real world rarely follow a standard normal distribution, z-scores are convenient calculations of area that can be used with any/all normal distributions. Meaning: once a researcher has translated raw data into a standard normal distribution (z-score), he/she can then find its associated probability.
Wiki User
∙ 14y ago
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What is shapes of distribution?
It is any shape that you want, provided that the total area under the curve is 1.
How do you write 12000 in standard form?
This might not be what you want to know but: 12,000 in standard form is 12,000 12,000 in expanded for is 10,000+ 2,000 Which is 12,000 in standard form
You want 9th standard maths solved model papers?
trignometry
What is a giga in standard form?
If you want to write Giga in standard form you put it as x109. So if you wanted to convert 2GV to volts it would be 2x109 V.
Eight million eight thousand in standard form?
Eight million eight thousand in standard form is 8,800,000. If you want to write that in scientific notation, then you would say 8.8 x 106
Why does a researcher want to go from a normal distribution to a standard normal distributio?
A normal distribution simply enables you to convert your values, which are in some measurement unit, to normal deviates. Normal deviates (i.e. z-scores) allow you to use the table of normal values to compute probabilities under the normal curve.
Would you want your grades based on normal distribution why or why not?
While the initial standard should be based on a distribution of some kind, and possibly not even the normal, each personal grade should not be based on it.
What is paramtric and non-paramtreic?
In statistics, an underlying assumption of parametric tests or analyses is that the dataset on which you want to use the test has been demonstrated to have a normal distribution. That is, estimation of the "parameters", such as mean and standard deviation, is meaningful. For instance you can calculate the standard deviation of any dataset, but it only accurately describes the distribution of values around the mean if you have a normal distribution. If you can't demonstrate that your sample is normally distributed, you have to use non-parametric tests on your dataset.
Why the normal distribution is important in public health research?
I just want to have the question answered for class
How do you calculate a coefficient knowing the expected rate of return and standard deviation?
It depends on what the underlying distribution is and which coefficient you want to calculate.
How many standard deviation tend to cover the entire range of scores in a distribution?
Type your answer here... It depends what percentage of the total data you want to embrace. 99.73% of the total distribution lies between minus to plus 3 standard deviations. That's usually the benchmark range.
What is the probability of P z-0.41 using the standard normal distribution?
See the related link for the area at 0.41 (same as -0.41) which is 0.1591. This area, which is the probability, is from minus infinity to -0.41. If you want the area from -0.41 to plus infinity you need to take 1 - 0.1591 which is 0.8409.
What test to use for non-normal distribution with data that's not nominal?
It depends on what you want to test. Goodnesss of fit or some null hypothesis?
Scores on the sat exam approximate a normal distribution with µ equals 500 and sd equals 100 use this distribution to determine the percentage of sat scores that fall above 600?
The answer is about 16% Using the z-score formula(z = (x-u)/sd) the z score is 1. This means that we want the percentage above 1 standard deviation. We know from the 68-95-99.7 rule that 68 percent of all the data fall between -1 and 1 standard deviation so there must be about 16% that falls above 1 standard deviation.
What is this 0.3989?
The equation for the normal distribution is given by:P(x) = 1/(σ√(2π)) * e^(-(x-µ)²/(2σ²))If we want to find P(x) maximum when µ = 0 and σ =1, then we substitute x =0. Giving:P(0) = 1/(1√(2π)) * e^(-(0-0)²/(2(1)²))= 1/(√(2π))≈ 0.3989This is also the value of the standard deviation if you were to produce a normal distribution with a maximum of 1. In order to find this σ, you must solve σ in P(x) = 1/(σ√(2π)) * e^(-(x-µ)²/(2σ²)) = 1. As the largest value that e^(-(x-µ)²/(2σ²)) can take is 1, (when (-(x-µ)²/(2σ²)) = 0), you can solve σ in:1/(σ√(2π)) = 1σ = 1/(√(2π))
For a normal distribution find the z-scores values that separate the middle 60 percent of the distribution for the 40 percent in the tails?
The standard normal table tells us the area under a normal curve to the left of a number z. The tables usually give only the positive value since one can use symmetry to find the corresponding negative values. The middle 60 percent leaves 20 percent on either side. So we want the z scores that correspond to that 80 percentile which is .804. Therefore the values are are between z scores of -.804 and .804 * * * * * I make it -0.8416 to 0.8416
Should you use the Binomial Normal or Poisson distribution if in the past few years an average of 10 businesses closed and I want to find the probability of more than 10 businesses closing next year?
If this is the only information that you have then you must use the Poisson distribution.
