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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

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.

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.

Formula review

The Central Limit Theorem for Sums: ∑X ~ N [( n )( μ x ),( n )( σ x )]

Mean for Sums ( ∑X ): ( n )( μ x )

The Central Limit Theorem for Sums z -score and standard deviation for sums: z for the sample mean =  Σ x ( n ) ( μ X ) ( n ) ( σ X )

Standard deviation for Sums ( ∑X ): ( n ) ( σ x )

Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

Find the probability that the sum of the 95 values is greater than 7,650.

0.3345

Find the probability that the sum of the 95 values is less than 7,400.

Find the sum that is two standard deviations above the mean of the sums.

7,833.92

Find the sum that is 1.5 standard deviations below the mean of the sums.


Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

Find the probability that the sum of the 40 values is greater than 7,500.

0.0089

Find the probability that the sum of the 40 values is less than 7,000.

Find the sum that is one standard deviation above the mean of the sums.

7,326.49

Find the sum that is 1.5 standard deviations below the mean of the sums.

Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

77.45%


Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.

Find the probability that the sum of the 100 values is greater than 3,910.

Find the probability that the sum of the 100 values is less than 3,900.

0.4207

Find the probability that the sum of the 100 values falls between the numbers you found in [link] and [link] .

Find the sum with a z –score of –2.5.

3,888.5

Find the sum with a z –score of 0.5.

Find the probability that the sums will fall between the z -scores –2 and 1.

0.8186


Use the following information to answer the next four exercise: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

What is the mean of ΣX ?

What is the standard deviation of ΣX ?

5

What is P ( Σx = 290)?

What is P ( Σx >290)?

0.9772

True or False: only the sums of normal distributions are also normal distributions.

In order for the sums of a distribution to approach a normal distribution, what must be true?

The sample size, n , gets larger.

What three things must you know about a distribution to find the probability of sums?

An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one object from this distribution. What is the sample size if the standard deviation of ΣX is 42?

49

An unknown distribution has a mean of 19 and a standard deviation of 20. Let X = the object of interest. What is the sample size if the mean of ΣX is 15,200?


Use the following information to answer the next three exercises.
A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers.

What is the z -score for Σx = 840?

26.00

What is the z -score for Σx = 1,186?

What is P ( Σx <1,186)?

0.1587


Use the following information to answer the next three exercises:
An unkwon distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one object of interest.

What is the mean of ΣX ?

What is the standard deviation of ΣX ?

1,000

What is P ( Σx >9,000)?

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Source:  OpenStax, Statistics i - math1020 - red river college - version 2015 revision a - draft 2015-10-24. OpenStax CNX. Oct 24, 2015 Download for free at http://legacy.cnx.org/content/col11891/1.8
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