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Sagot :
The percentage of the class that passed is 16%.
Since we require at least 50% to pass the test and the mean of test is X = 46 % and its standard deviation is σ = 4%, we need to find how many standard deviations away from the mean contain 50% or more.
Now, everything below 46% which is the mean failed and this is 50 % of the class, if the scores were normally distributed,
Number of standard deviations away from the mean
We need to find how many standard deviations away from the mean is 50 %.
So, n = (X - μ)/σ where
- X = score = 50 %,
- μ = mean = 46 % and
- σ = standard deviation = 4 %
So, n = (X - μ)/σ
n = (50% - 46%)/4%
n = 4%/4%
n = 1
So, at 1 standard deviation away from the mean, there are 34% of the values contanined.
Percentage of class who failed the test
So, the percentage of the class less than 50% is 50% + 34% = 84 %.
Percentage of class who passed the test
The percentage of the class greater than 50% is 100% - 84% = 16%
So, the percentage of the class that passed is 16%.
Learn more about normal distribution here:
https://brainly.com/question/16943251
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