Respuesta :
Answer:
The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.
Step-by-step explanation:
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
[tex]\mu_{\bar x}=\mu[/tex]
And the standard deviation of the distribution of sample mean is given by,
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}[/tex]
The information provided is:
μ = 144 mm
σ = 7 mm
n = 50.
Since n = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.
[tex]\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)[/tex]
Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:
[tex]P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})[/tex]
[tex]=P(Z>2.63)\\=1-P(Z<2.63)\\=1-0.99573\\=0.00427\\\approx 0.0043[/tex]
*Use a z-table for the probability.
Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.
