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In a recent study reported Oct.29, 2012 on the Flurry Blog, the mean age of tablet users is 35 years. Suppose the standard deviation is ten years. The sample size is 39.

  1. What are the mean and standard deviation for the sum of the ages of tablet users? What is the distribution?
  2. Find the probability that the sum of the ages is between 1,400 and 1,500 years.
  3. Find the 90 th percentile for the sum of the 39 ages.
  1. μ Σx = x = 1,365 and σ Σx = n σ x = 62.4
    The distribution is normal for sums by the central limit theorem.
  2. P (1400< Σ x <1500) = normalcdf (1400,1500,(39)(35),( 39 )(10)) = 0.2723
  3. Let k = the 90 th percentile.
    k  =  invNorm (0.90,(39)(35),( 39 ) (10)) = 1445.0
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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.

  1. What are the mean and standard deviation for the sums?
  2. Find the 95 th percentile for the sum of the sample. Interpret this value in a complete sentence.
  3. Find the probability that the sum of the sample is at least ten hours.
  1. μ Σx = x = 70(8.2) = 574 minutes and σ Σx = ( n ) ( σ x ) = ( 70  ) (1) = 8.37 minutes
  2. Let k = the 95 th percentile.
    k = invNorm (0.95,(70)(8.2), ( 70 ) (1)) = 587.76 minutes
    Ninety five percent of the app engagement times are at most 587.76 minutes.
  3. ten hours = 600 minutes
    P x ≥ 600) = normalcdf (600,E99,(70)(8.2), ( 70 ) (1)) = 0.0009
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The mean number of minutes for app engagement by a table use is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample size of 70.

  1. What is the probability that the sum of the sample is between seven hours and ten hours? What does this mean in context of the problem?
  2. Find the 84 th and 16 th percentiles for the sum of the sample. Interpret these values in context.
  1. 7 hours = 420 minutes
    10 hours = 600 minutes
    normalcdf P ( 420 Σ x 600 ) = n o r m a l c d f ( 420 , 600 , ( 70 ) ( 8.2 ) , 70 ( 1 ) ) = 0.9991
    This means that for this sample sums there is a 99.9% chance that the sums of usage minutes will be between 420 minutes and 600 minutes.
  2. i n v N o r m ( 0.84 , ( 70 ) ( 8.2 ) , 70 ( 1 ) ) = 582.32
    i n v N o r m ( 0.16 , ( 70 ) ( 8.2 ) , 70 ( 1 ) ) = 565.68
    Since 84% of the app engagement times are at most 582.32 minutes and 16% of the app engagement times are at most 565.68 minutes, we may state that 68% of the app engagement times are between 565.68 minutes and 582.32 minutes.
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References

Farago, Peter. “The Truth About Cats and Dogs: Smartphone vs Tablet Usage Differences.” The Flurry Blog, 2013. Posted October 29, 2012. Available online at http://blog.flurry.com (accessed May 17, 2013).

Chapter review

The central limit theorem tells us that for a population with any distribution, the distribution of the sums for the sample means approaches a normal distribution as the sample size increases. In other words, if the sample size is large enough, the distribution of the sums can be approximated by a normal distribution even if the original population is not normally distributed. Additionally, if the original population has a mean of μ X and a standard deviation of σ x , the mean of the sums is x and the standard deviation is ( n ) ( σ x ) where n is the sample size.

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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