Respuesta :
Answer:
The probability that the proportion of passed keypads is between 0.72 and 0.80 is 0.6677.
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
Let p = the proportion of keypads that pass inspection at a cell phone assembly plant.
The probability that a randomly selected cell phone keypad passes the inspection is, p = 0.77.
A random sample of n = 111 keypads is analyzed.
Then the sampling distribution of [tex]\hat p[/tex] is:
[tex]\hat p\sim N(p,\ \sqrt{\frac{p(1-p)}{n}})\\\Rightarrow \hat p\sim N (0.77, 0.04)[/tex]
Compute the probability that the proportion of passed keypads is between 0.72 and 0.80 as follows:
[tex]P(0.72<\hat p<0.80)=P(\frac{0.72-0.77}{0.04}<\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.80-0.77}{0.04})[/tex]
[tex]=P(-1.25<Z<0.75)\\=P(Z<0.75)-P(Z<-1.25)\\=0.77337-0.10565\\=0.66772\\\approx0.6677[/tex]
Thus, the probability that the proportion of passed keypads is between 0.72 and 0.80 is 0.6677.
Using the normal distribution and the central limit theorem, it is found that there is a 0.6677 = 66.77% probability that the proportion of the sample keypads that pass inspection is between 0.72 and 0.8.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample proportions of a proportion p in a sample of size n has mean [tex]\mu = p[/tex] and standard error [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex].
In this problem:
- 77% of the cell phone keypads pass inspection, hence [tex]p = 0.77[/tex].
- A random sample of 111 keypads is analyzed, hence [tex]n = 111[/tex].
The mean and the standard error are given by:
[tex]\mu = p = 0.77[/tex]
[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.77(0.23)}{111}} = 0.0399[/tex]
The probability is the p-value of Z when X = 0.8 subtracted by the p-value of Z when X = 0.72.
X = 0.8:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.8 - 0.77}{0.0399}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a p-value of 0.7734.
X = 0.72:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.72 - 0.77}{0.0399}[/tex]
[tex]Z = -1.25[/tex]
[tex]Z = -1.25[/tex] has a p-value of 0.1057.
0.7734 - 0.1057 = 0.6677
0.6677 = 66.77% probability that the proportion of the sample keypads that pass inspection is between 0.72 and 0.8.
For more on the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213
