Respuesta :
Answer:
75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32
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.
For the sampling distribution of a proportion p with size n, we have that [tex]\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this problem, we have that:
[tex]p = 0.3, n= = 245[/tex]
So
[tex]\mu = 0.3, \sigma = \sqrt{\frac{0.3*0.7}{245}} = 0.0293[/tex]
What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32
This is the pvalue of Z when X = 0.32. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.32 - 0.3}{0.0293}[/tex]
[tex]Z = 0.68[/tex]
[tex]Z = 0.68[/tex] has a pvalue of 0.7517
75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32
