Respuesta :
Answer:
0.2088 = 20.88% probability that Jackson randomly selects a person from this population who recalls between 26 and 29 memories.
Step-by-step explanation:
Normal Probability Distribution:
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Population mean of 30 and a population standard deviation of 5.
This means that [tex]\mu = 30, \sigma = 5[/tex]
What is the probability that Jackson randomly selects a person from this population who recalls between 26 and 29 memories?
This is the pvalue of Z when X = 29 subtracted by the pvalue of Z when X = 26. So
X = 29
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{29 - 30}{5}[/tex]
[tex]Z = -0.2[/tex]
[tex]Z = -0.2[/tex] has a pvalue of 0.4207
X = 26
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{26 - 30}{5}[/tex]
[tex]Z = -0.8[/tex]
[tex]Z = -0.8[/tex] has a pvalue of 0.2119
0.4207 - 0.2119 = 0.2088
0.2088 = 20.88% probability that Jackson randomly selects a person from this population who recalls between 26 and 29 memories.
