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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:
If you draw random samples of size n , then as n increases, the random variable Σ X consisting of sums tends to be normally distributed and Σ Χ ~ N (( n )( μ Χ ), ( )( σ Χ )).
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 deviation equal 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:
To find probabilities for sums on the calculator, follow these steps.
2
nd
DISTR
2:
normalcdf
normalcdf
(lower value of the area, upper value of the area, (
n )(mean), (
)(standard deviation))
where:
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.
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.
Σ
X = the sum or total of 80 values. Since
μ
X = 90,
σ
X = 15, and
n = 80,
~
N ((80)(90),
(
)(15))
a. Find P (Σ x >7,500)
P (Σ x >7,500) = 0.0127
normalcdf
(lower value, upper value, mean of sums,
stdev
of sums)
The parameter list is abbreviated(lower, upper, ( n )( μ X , ( σ X ))
normalcdf
(7500,1E99,(80)(90),
(15)) = 0.0127
1E99 = 10 99 .
Press the
EE
key for E.
b. Find Σ x where z = 1.5.
Σ x = ( n )( μ X ) + ( z ) ( σ Χ ) = (80)(90) + (1.5)( )(15) = 7,401.2
An unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.
0.0040
To find percentiles for sums on the calculator, follow these steps.
2
nd DIStR
3:invNorm
k = invNorm (area to the left of
k , (
n )(mean),
(standard deviation))
where:
In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50.
normalcdf
(1,500, 1,800, (50)(34),
(15)) = 0.7974invNorm
(0.80,(50)(34),
(15)) = 1,789.3Notification Switch
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