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

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

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Vodka 5 days 2021-07-22T16:13:22+00:00 1 Answers 2 views 0

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    2021-07-22T16:14:47+00:00

    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

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