SOLUTION: A worldwide organization of academics claims that the mean IQ score of its members is 115
, with a standard deviation of 16
. A randomly selected group of 40
members of this or
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-> SOLUTION: A worldwide organization of academics claims that the mean IQ score of its members is 115
, with a standard deviation of 16
. A randomly selected group of 40
members of this or
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Question 1135014: A worldwide organization of academics claims that the mean IQ score of its members is 115
, with a standard deviation of 16
. A randomly selected group of 40
members of this organization is tested, and the results reveal that the mean IQ score in this sample is 115.8
. If the organization's claim is correct, what is the probability of having a sample mean of 115.8
or less for a random sample of this size? Answer by Theo(13342) (Show Source):
z is the z-score
x is the raw score
m is the mean
s is the standard error.
the formula for standard error is s = sd / sqrt(n)
sd is the standard deviation of the population.
sqrt(n) is the square root of the sample size.
the claim is that the population has a mean of 115 with a standard deviation of 16.
the sample of 40 members has a mean of 115.8.
the standard error for the distribuion of sample means is equal to 16 / sqrt(40) which is equal to 2.529822128.
the z-score for the sample mean compared to the population mean would be equal to (115.8 - 115) / 2.529822128 which is equal to .316227766.
the probability of having a sample mean of 15.8 or lessfor a random sample of this size would be the area under the normal distribution curve to the left of a z-score of .316227766.
using the z-score calculator on the TI-84 Plus, i get this area to be equal to .6240851183.
the probability of getting a sample of size 40 with a mean less than 115.8 is approximately 62.41%.
this can be seen visually through the use of the following online normal distribution calculator.
