Respuesta :
Answer:
0.9599
Step-by-step explanation:
This is a normal distribution problem with
Mean = μ = 55 mm
Standard deviation = σ = 4 mm
The probability that the diameter of a selected bearing is greater than 48 millimeters.
We first standardize the bearing diameter, 48 mm
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (48 - 55)/4 = - 1.75
To determine the probability that the diameter of a selected bearing is greater than 48 millimeters.
P(x > 48) = P(z > -1.75)
We'll use data from the normal probability table for these probabilities
P(x > 48) = P(z > -1.75) = 1 - P(z ≤ -1.75) = 1 - 0.04006 = 0.95994 = 0.9599 to 4 d.p
Hope this Helps!!!
Answer:
[tex]P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-55}{4})=P(z>-1.75)[/tex]
And we can find this probability using the complement rule and we got:
[tex]P(z>-1.75)=1-P(z<-1.75)=1-0.04= 0.96[/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 diameters of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(55,4)[/tex]
Where [tex]\mu=55[/tex] and [tex]\sigma=4[/tex]
We are interested on this probability
[tex]P(X>48)[/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]
If we apply this formula to our probability we got this:
[tex]P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-55}{4})=P(z>-1.75)[/tex]
And we can find this probability using the complement rule and we got:
[tex]P(z>-1.75)=1-P(z<-1.75)=1-0.04= 0.96[/tex]