Respuesta :
Using the normal distribution, there is a 0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].
The estimate and the sample size are:
p = 0.29, n = 207.
Hence the mean and the standard error are given as follows:
- [tex]\mu = p = 0.29[/tex].
- [tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.29(0.71)}{207}} = 0.0315[/tex].
The probability that the sample proportion of households spending more than $125 a week is less than 0.31 is the p-value of Z when X = 0.31, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.31 - 0.29}{0.0315}[/tex]
Z = 0.63
Z = 0.63 has a p-value of 0.7357.
0.7357 = 73.57% probability that the sample proportion of households spending more than $125 a week is less than 0.31.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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