What proportion of the students scored at least 23 points on this test, rounded to five decimal places

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?

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

1 is c

2 is a

3 is a

step-by-step explanation:

answer is : muchas/tipicas

have a nice day!

~recklesschicken