This question is incomplete, the complete question is;
The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, DO NOT apply the standard deviation rule.
What proportion of the students scored at least 23 points on this test, rounded to five decimal places?
Answer:
proportion of the students that scored at least 23 points on this test is 0.30850
Step-by-step explanation:
Given the data in the question;
mean μ = 22
standard deviation σ = 2
since test closely followed a Normal Distribution
let
Z = x-μ / σ { standard normal random variable ]
Now, proportion of the students that scored at least 23 points on this test.
P( x ≥ 23 ) = P( (x-μ / σ) ≥ ( 23-22 / 2 )
= P( Z ≥ 1/2 )
= P( Z ≥ 0.5 )
= 1 - P( Z < 0.5 )
Now, from z table
{ we have P( Z < 0.5 ) = 0.6915 }
= 1 - P( Z < 0.5 ) = 1 - 0.6915 = 0.30850
P( x ≥ 23 ) = 0.30850
Therefore, proportion of the students that scored at least 23 points on this test is 0.30850
