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Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:

  • μ X = the mean of X
  • σ X = the standard deviation of X
If you draw random samples of size n , then as n increases, the random variable ΣX which consists of sums tends to be normally distributed and

Σ X ~ N ( n μ X , n σ X )

The Central Limit Theorem for Sums says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution) which approaches a normal distribution as the sample size increases. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviationequal to the original standard deviation multiplied by the square root of the sample size.

The random variable Σ X has the following z-score associated with it:

  • Σx is one sum.
  • z = Σ x - n μ X n σ X
  • n μ X = the mean of ΣX
  • n σ X = standard deviation of ΣX

An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.

  • Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7500.
  • Find the sum that is 1.5 standard deviations above the mean of the sums.

Let X = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.

ΣX = the sum or total of 80 values. Since μ X = 90 , σ X = 15 , and n = 80 , then

Σ X ~ N ( 80 90 , 80 15 )

  • mean of the sums = n μ X = ( 80 ) ( 90 ) = 7200
  • standard deviation of the sums = n σ X = 80 15
  • sum of 80 values = Σx = 7500

  • Find P ( Σx 7500 )

P ( Σx 7500 ) = 0.0127

Normal distribution curve of sum X with the values of 7200 and 7500 on the x-axis. A vertical upward line extends from point 7500 on the x-axis up to the curve. The probability area occurs from point 7500 and to the end of the curve.

normalcdf (lower value, upper value, mean of sums, stdev of sums)

The parameter list is abbreviated (lower, upper, n μ X , n σ X )

normalcdf (7500,1E99, 80 90 , 80 15 ) = 0.0127

Reminder: 1E99 = 10 99 . Press the EE key for E.

  • Find Σx where z = 1.5:


Σx = n μ X + z n σ X = (80)(90) + (1.5)( 80 ) (15)= 7401.2

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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Abhi
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ninjadapaul
20/(×-6^2)
Salomon
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ninjadapaul
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ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
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Abhi
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Collaborative statistics (custom lecture version modified by t. short). OpenStax CNX. Jul 15, 2013 Download for free at http://cnx.org/content/col11543/1.1
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