Respuesta :
Answer:
2.89% probability that weight for a part made using the new production process is less than 1.61 or greater than 2.53 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 2.09, \sigma = 0.21[/tex]
Find the probability that weight for a part made using the new production process is less than 1.61 or greater than 2.53 pounds
Less than 1.61
This is the pvalue of Z when X = 1.61. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1.61 - 2.09}{0.21}[/tex]
[tex]Z = -2.29[/tex]
[tex]Z = -2.29[/tex] has a pvalue of 0.0110
Greater than 2.53
1 subtracted by the pvalue of Z when X = 2.53.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2.53 - 2.09}{0.21}[/tex]
[tex]Z = 2.1[/tex]
[tex]Z = 2.1[/tex] has a pvalue of 0.9821
1 - 0.9821 = 0.0179
Less than 1.61 or greater than 2.53 pounds
0.0110 + 0.0179 = 0.0289
2.89% probability that weight for a part made using the new production process is less than 1.61 or greater than 2.53 pounds
Answer:
[tex] P(X<1.61) = P(z<\frac{1.61-2.09}{0.21}) = P(Z<-2.286)=0.011[/tex]
[tex] P(X>2.53) = P(z>\frac{2.53-2.09}{0.21}) = P(Z>2.095)=1-P(Z<2.095)= 1-0.982= 0.018[/tex]
And adding the probabilites we got:
[tex]P(X< 1.61 \cup X>2.53)= 0.011+ 0.018 = 0.029[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(2.09,0.21)[/tex]
Where [tex]\mu=2.09[/tex] and [tex]\sigma=0.21[/tex]
We are interested on this probability
[tex]P(X< 1.61 \cup X>2.53)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
We can find the individual probabilites like this:
[tex] P(X<1.61) = P(z<\frac{1.61-2.09}{0.21}) = P(Z<-2.286)=0.011[/tex]
[tex] P(X>2.53) = P(z>\frac{2.53-2.09}{0.21}) = P(Z>2.095)=1-P(Z<2.095)= 1-0.982= 0.018[/tex]
And adding the probabilites we got:
[tex]P(X< 1.61 \cup X>2.53)= 0.011+ 0.018 = 0.029[/tex]
