Respuesta :
Answer:
51.60% probability that a randomly selected adult has an IQ between 86 and 114.
Step-by-step explanation:
Problems of normally distributed samples are 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 problem, we have that:
[tex]\mu = 114, \sigma = 86[/tex]
Find the probability that a randomly selected adult has an IQ between 86 and 114.
Pvalue of Z when X = 114 subtracted by the pvalue of Z when X = 86. So
X = 114
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{114 - 100}{20}[/tex]
[tex]Z = 0.7[/tex]
[tex]Z = 0.7[/tex] has a pvalue of 0.7580
X = 86
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{86 - 100}{20}[/tex]
[tex]Z = -0.7[/tex]
[tex]Z = -0.7[/tex] has a pvalue of 0.2420
0.7580 - 0.2420 = 0.5160
51.60% probability that a randomly selected adult has an IQ between 86 and 114.
