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Question 1161981: The average grade of a subject is 1.89 for the whole class, with a standard deviation of 0.63. If a sample is 25 students is being taken, then find the probability of getting the average of this sample to be within the range of 1.6 and 2.0.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! z = (x - m) / s
z is the z-score
x is the raw score
m is the raw mean
n is the sample size
s is the standard deviation of the distribution of sample means
sd is the standard deviation of the population, or the sample, if the standard deviation of the population is not available.
if the standard deviation of the population is known, use the z-score, else use the t-score.
in your problem:
m = 1.89
sd = .63
n = 25
s = sd / sqrt(n) = .63 / sqrt(25) = .126
use the z-score formula to find the z-scores.
that formula is z = (x - m) / s
for a raw score of 1.6, that formula becomes:
z = (1.6 - 1.89) / .126 = -2.30 rounded to 2 decimal places.
for a raw score of 2.0, tha formula becomes:
z = (2.0 - 1.89) / .126 = .87 rounded to 2 decimal places.
look for the area to the left of the z-score equal to -2.30 and .87 in the z-score table.
you will find that the area to the left of each z-score is as indicated below:
for -2.30 = .01072
for .87 = .80785
subtract the smaller area from the larger area to get the areas in between is equal to .79713.
that's the probability that the scores in this sample will be between 1.6 and 2.0.
the table i used can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
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