Respuesta :
Answer:
75.94% of the households spent between $5.00 and $9.00 on sugar.
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 = 8.22, \sigma = 1.10[/tex]
What proportion of the households spent between $5.00 and $9.00 on sugar?
This is the pvalue of Z when X = 9 subtracted by the pvalue of Z when X = 5. So
X = 9
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9 - 8.22}{1.10}[/tex]
[tex]Z = 0.71[/tex]
[tex]Z = 0.71[/tex] has a pvalue of 0.7611
X = 5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 8.22}{1.10}[/tex]
[tex]Z = -2.93[/tex]
[tex]Z = -2.93[/tex] has a pvalue of 0.0017.
So 0.7611 - 0.0017 = 0.7594 = 75.94% of the households spent between $5.00 and $9.00 on sugar.
