The scores on a certain test are normally distributed with a mean score of 70 and a standard deviation of 3. What is the probability that a sample of 90 students will have a mean score of at least 70.3162?
Mean = = 70
Standard deviation = = 3
Sample size = n = 90
We have to find P( 70.3162)
For finding this probability we have to find z score.
That is we have to find P(Z 1)
P(Z 1)= 1 - P(Z < 1) = 1 - 0.8413 = 0.1587
( From z table)
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