Respuesta :
Answer:
57.62% probability that a randomly selected adult has an IQ between 89 and 121.
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, we have that:
[tex]\mu = 105, \sigma = 20[/tex]
Find the probability that a randomly selected adult has an IQ between 89 and 121.
This is the pvalue of Z when X = 121 subtracted by the pvalue of Z when X = 89. So
X = 121
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{121 - 105}{20}[/tex]
[tex]Z = 0.8[/tex]
[tex]Z = 0.8[/tex] has a pvalue of 0.7881
X = 89
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{89 - 105}{20}[/tex]
[tex]Z = -0.8[/tex]
[tex]Z = -0.8[/tex] has a pvalue of 0.2119
0.7881 - 0.2119 = 0.5762
57.62% probability that a randomly selected adult has an IQ between 89 and 121.
