Respuesta :
Answer:
96.4% of people have an IQ score between 69 and 132
Step-by-step explanation:
When the distribution is normal, we use 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:
[tex]\mu = 100, \sigma = 15[/tex]
What percentage of people have an IQ score between 69 and 132, to the nearest tenth?
The proportion is the pvalue of Z when X = 142 subtracted by the pvalue of Z when X = 69. The percentage is the proportion multiplied by 100. So
X = 132
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{132 - 100}{15}[/tex]
[tex]Z = 2.13[/tex]
[tex]Z = 2.13[/tex] has a pvalue of 0.983
X = 69
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{69 - 100}{15}[/tex]
[tex]Z = -2.07[/tex]
[tex]Z = -2.07[/tex] has a pvalue of 0.019
0.983 - 0.019 = 0.964
0.964*100 = 96.4%
96.4% of people have an IQ score between 69 and 132
