Respuesta :
Answer:
81.85% of the respondents will be between 64.29 and 93.54 years old
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 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.
Mean and standard deviation are 74.04 and 9.75
This means that [tex]\mu = 74.04, \sigma = 9.75[/tex]
What percentage of the respondents will be between 64.29 and 93.54 years old?
The proportion is the pvalue of Z when X = 93.54 subtracted by the pvalue of Z when X = 64.29. So
X = 93.54
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{93.54 - 74.04}{9.75}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772.
X = 64.29
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{64.29 - 74.04}{9.75}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a pvalue of 0.1587.
0.9772 - 0.1587 = 0.8185
0.8185*100 = 81.85%
81.85% of the respondents will be between 64.29 and 93.54 years old
