. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person selected at

Question

. If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person selected at random will have an IQ of 110 or greater?

in progress 0
Claire 1 week 2021-11-25T14:12:47+00:00 1 Answer 0 views 0

Answers ( )

    0
    2021-11-25T14:13:52+00:00

    Answer:

    2.28% probability that a person selected at random will have an IQ of 110 or greater

    Step-by-step explanation:

    Problems of normally distributed samples are solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    In this problem, we have that:

    \mu = 100, \sigma = 5

    What is the probability that a person selected at random will have an IQ of 110 or greater?

    This is 1 subtracted by the pvalue of Z when X = 110. So

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{110 - 100}{5}

    Z = 2

    Z = 2 has a pvalue of 0.9772

    1 – 0.9772 = 0.0228

    2.28% probability that a person selected at random will have an IQ of 110 or greater

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )