Scores on a certain intelligence test for children between ages 13 and 15 years are approximately normally distributed with μ=101 and σ=24. (a) what proportion of children aged 13 to 15 years old have scores on this test above 83 ? (note: please enter your answer in decimal form. for example, 45.23% should be entered as 0.4523.)

Answer:

a) 0.7734

Step-by-step explanation:

Problems of normal distributions can be solved using the z-score formula.

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 question, we have that:

[tex]\mu = 101, \sigma = 24[/tex]

(a) what proportion of children aged 13 to 15 years old have scores on this test above 83 ?

This is 1 subtracted by the pvalue of Z when X = 83. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{83 - 101}{24}[/tex]

[tex]Z = -0.75[/tex]

[tex]Z = -0.75[/tex] has a pvalue of 0.2266

1 - 0.2266 = 0.7734

The answer is 0.7734

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