Respuesta :
Answer:
40.65% of these mothers are between the ages of 32 to 40
23.27% of these mothers are less than 30 years old
37.07% of these mothers are more than 38 years old
Step-by-step explanation:
Problems of normally distributed samples 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.
In this problem, we have that:
[tex]\mu = 35.5, \sigma = 7.5[/tex]
What percent of these mothers are between the ages of 32 to 40?
This is the pvalue of Z when X = 40 subtracted by the pvalue of Z when X = 32.
X = 40
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{40 - 35.5}{7.5}[/tex]
[tex]Z = 0.6[/tex]
[tex]Z = 0.6[/tex] has a pvalue of 0.7257
X = 32
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32 - 35.5}{7.5}[/tex]
[tex]Z = -0.47[/tex]
[tex]Z = -0.47[/tex] has a pvalue of 0.3192
0.7257 - 0.3192 = 0.4065
40.65% of these mothers are between the ages of 32 to 40
What percent of these mothers are less than 30 years old?
This is the pvalue of Z when X = 30.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30 - 35.5}{7.5}[/tex]
[tex]Z = -0.73[/tex]
[tex]Z = -0.73[/tex] has a pvalue of 0.2327
23.27% of these mothers are less than 30 years old
What percent of these mothers are more than 38 years old?
This is 1 subtracted by the pvalue of Z when X = 38.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{38 - 35.5}{7.5}[/tex]
[tex]Z = 0.33[/tex]
[tex]Z = 0.33[/tex] has a pvalue of 0.6293
1 - 0.6293 = 0.3707
37.07% of these mothers are more than 38 years old
