Respuesta :
Answer:
The probability that the sample mean would differ from the population mean by more than 0.5 millimeters is 0.2420.
Step-by-step explanation:
Let X = the diameter of the steel bolts manufactured by the steel bolts manufacturing company Thompson and Thompson.
The mean diameter of the bolts is:
μ = 149 mm.
The standard deviation of the diameter of bolts is:
σ = 5 mm.
A random sample, of size n = 49, of steel bolts are selected.
The population of the diameter of bolts is not known.
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
The mean of the sampling distribution of sample mean is:
[tex]\mu_{\bar x}=\mu=149\ mm[/tex]
The standard deviation of the sampling distribution of sample mean is:
[tex]\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{49}}=0.7143[/tex]
Compute the probability that the sample mean would differ from the population mean by more than 0.5 millimeters as follows:
[tex]P(\bar X>\bar x)=P(\frac{\bar X-\mu_{\bar x}}{\sigma _{\bar x}}>\frac{0.50}{0.7143})\\=P(Z>0.70)\\=1-P(Z<0.70)\\=1-0.7580\\=0.2420[/tex]
*Use a z-table for the probability.
Thus, the probability that the sample mean would differ from the population mean by more than 0.5 millimeters is 0.2420.
Answer: The required probability is 0.4839.
Step-by-step explanation:
Since we have given that
Mean = 149 millimeter
Standard deviation = 5 millimeter
n = 49
So, we have , [tex]\mu_x=149\ and\ \sigma_x=\dfrac{5}{\sqrt{49}}=\dfrac{5}{7}=0.714[/tex]
So, it becomes,
[tex]1-P(-0.5>X>0.5)\\\\=1-P(\dfrac{-0.5}{0.71}>z>\dfrac{0.5}{0.71})\\\\=1-P(-0.70>z>0.70)\\\\=1-P(z<-0.70)-P(z<0.7)\\\\=1-(0.24196+0.75804)\\\\=1-0.51608\\\\=0.48392[/tex]
Hence, the required probability is 0.4839.
