Question

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 149 millimeters, and a standard deviation of 5 millimeters. If a random sample of 49 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by more than 0.5 millimeters? Round your answer to four decimal places.

Answer 1

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:

$\mu_{\bar x}=\mu=149\ mm$

The standard deviation of the sampling distribution of sample mean is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{49}}=0.7143$

Compute the probability that the sample mean would differ from the population mean by more than 0.5 millimeters as follows:

$P(\bar X>\bar x)=P(\frac{\bar X-\mu_{\bar x}}{\sigma _{\bar x}}>\frac{0.50}{0.7143})\\=P(Z>0.70)\\=1-P(Z$

*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 2

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 , $\mu_x=149\ and\ \sigma_x=\dfrac{5}{\sqrt{49}}=\dfrac{5}{7}=0.714$

So, it becomes,

$1-P(-0.5>X>0.5)\\\\=1-P(\dfrac{-0.5}{0.71}>z>\dfrac{0.5}{0.71})\\\\=1-P(-0.70>z>0.70)\\\\=1-P(z$

Hence, the required probability is 0.4839.