Suppose X is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:
μX = the mean of Χ
σΧ = the standard deviation of X
If you draw random samples of size n, then as n increases, the random variable ΣX consisting of sums tends to be normally distributed and ΣΧ ~ N((n)(μΧ), (n n)(σΧ)).
The central limit theorem for sums says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate the sum of each sample, these sums tend to follow a normal distribution. As sample sizes increase, the distribution of means more closely follows the normal distribution. 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 ΣX has the following z-score associated with it:
Σx is one sum.
z = Σx–(n)( μ X ) ( n )( σ X ) z = Σx–(n)( μ X ) ( n )( σ X ) (n)(μX) = the mean of ΣX ( n )( σ X ) ( n )( σ X ) = standard deviation of ΣX ΣX
Using the TI-83, 83+, 84, 84+ Calculator
To find probabilities for sums on the calculator, follow these steps.
2nd DISTR 2:normalcdf normalcdf(lower value of the area, upper value of the area, (n)(mean), (n n)(standard deviation))
where:
mean is the mean of the original distribution
standard deviation is the standard deviation of the original distribution
sample size = n
Example 7.5
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.
Problem
Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.
Find the sum that is 1.5 standard deviations above the mean of the sums.
Solution
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 μX = 90, σX = 15, and n = 80, ΣX ΣX ~ N((80)(90), (80 80)(15))
mean of the sums = (n)(μX) = (80)(90) = 7,200
standard deviation of the sums = ( n )( σ X ) = ( 80 ) ( n )( σ X ) = ( 80 )(15)
sum of 80 values = Σx = 7,500
a. Find P(Σx > 7,500)
P(Σx > 7,500) = 0.0127
Using the TI-83, 83+, 84, 84+ Calculator
normalcdf(lower value, upper value, mean of sums, stdev of sums)
The parameter list is abbreviated(lower, upper, (n)(μX, ( n ) ( n )(σX))
normalcdf (7500,1E99,(80)(90),( 80 ) ( 80 )(15)) = 0.0127
REMINDER
1E99 = 1099.
Press the EE key for E.
b. Find Σx where z = 1.5.
Σx = (n)(μX) + (z)( n ) ( n )(σΧ) = (80)(90) + (1.5)(80 80)(15) = 7,401.2
Try It 7.5
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.
Using the TI-83, 83+, 84, 84+ Calculator
To find percentiles for sums on the calculator, follow these steps.
2nd DIStR 3:invNorm k = invNorm (area to the left of k, (n)(mean), ( n ) ( n )(standard deviation))
where:
k is the kth percentile
mean is the mean of the original distribution
standard deviation is the standard deviation of the original distribution
sample size = n
Example 7.6
Problem
In a recent study, it was reported that the mean age of iPad users is 34 years. Suppose the standard deviation is 15 years. The sample of size is 50.
What are the mean and standard deviation for the sum of the ages of iPad users? What is the distribution?
Find the probability that the sum of the ages is between 1,500 and 1,800 years.
Find the 80th percentile for the sum of the 50 ages.
Solution
μΣx = nμx = 50(34) = 1,700 and σΣx = n nσx = ( 50 ) ( 50 )(15) = 106.07 The distribution is normal for sums by the central limit theorem.
P(1500 < Σx < 1800) = normalcdf (1,500, 1,800, (50)(34), ( 50 ) ( 50 )(15)) = 0.7974
Let k = the 80th percentile. k = invNorm(0.80,(50)(34),( 50 ) ( 50 )(15)) = 1,789.3
Try It 7.6
In a recent study, it was reported that the mean age of iPad users is 35 years. Suppose the standard deviation is ten years. The sample size is 39.
What are the mean and standard deviation for the sum of the ages of iPad users? What is the distribution?
Find the probability that the sum of the ages is between 1,400 and 1,500 years.
Find the 90th percentile for the sum of the 39 ages.
Example 7.7
Problem
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of size 70.
What are the mean and standard deviation for the sums?
Find the 95th percentile for the sum of the sample. Interpret this value in a complete sentence.
Find the probability that the sum of the sample is at least ten hours.
Solution
μΣx = nμx = 70(8.2) = 574 minutes and σΣx = ( n )( σ x ) ( n )( σ x ) = ( 70 ) ( 70 )(1) = 8.37 minutes
Let k = the 95th percentile. k = invNorm (0.95,(70)(8.2),( 70 ) ( 70 )(1)) = 587.76 minutes Ninety five percent of the sums of app engagement times are at most 587.76 minutes.
ten hours = 600 minutes P(Σx ≥ 600) = normalcdf(600,E99,(70)(8.2),( 70 ) ( 70 )(1)) = 0.0009
Try It 7.7
1. ਟੇਬਲ ਦੀ ਵਰਤੋਂ ਦੁਆਰਾ ਐਪ ਦੀ ਸ਼ਮੂਲੀਅਤ ਲਈ ਔਸਤ ਮਿੰਟਾਂ ਦੀ ਗਿਣਤੀ 8.2 ਮਿੰਟ ਹੈ। ਮੰਨ ਲਓ ਕਿ ਸਟੈਂਡਰਡ ਡਿਵੀਏਸ਼ਨ ਇੱਕ ਮਿੰਟ ਹੈ। ਨਮੂਨੇ ਦਾ ਆਕਾਰ 70 ਲਓ।
2. ਇਸ ਗੱਲ ਦੀ ਕੀ ਸੰਭਾਵਨਾ ਹੈ ਕਿ ਨਮੂਨੇ ਦਾ ਜੋੜ ਸੱਤ ਘੰਟਿਆਂ ਅਤੇ ਦਸ ਘੰਟਿਆਂ ਦੇ ਵਿਚਕਾਰ ਹੋਵੇ? ਇਸ ਸਮੱਸਿਆ ਦੇ ਸੰਦਰਭ ਵਿੱਚ ਇਸਦਾ ਕੀ ਮਤਲਬ ਹੈ?
3. ਨਮੂਨੇ ਦੇ ਜੋੜ ਲਈ 84ਵੇਂ ਅਤੇ 16ਵੇਂ ਪਰਸੈਂਟਾਈਲ ਲੱਭੋ। ਇਹਨਾਂ ਮੁੱਲਾਂ ਦੀ ਸੰਦਰਭ ਵਿੱਚ ਵਿਆਖਿਆ ਕਰੋ।