Respuesta :
Notice it is +/-4 times standard deviation (0.65-0.04 = 0.61, 0.69-0.04=0.65)
That's almost everything. So (D) is the answer.
99.993666%
That's almost everything. So (D) is the answer.
99.993666%
Answer:
D. 99.9936%.
Step-by-step explanation:
We have been given that the length of a screw produced by a machine is normally distributed with a mean of 0.65 inches and a standard deviation of 0.01 inches.
To find the percent of screws that are between 0.61 and 0.69 inches, first of we will find z-score for our given raw scores using z-score formula.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Raw-score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
Now let us find z-score corresponding to our given raw scores.
[tex]z=\frac{0.61-0.65}{0.01}[/tex]
[tex]z=\frac{-0.04}{0.01}[/tex]
[tex]z=-4[/tex]
Now let us find z-score for raw score 0.69.
[tex]z=\frac{0.69-0.65}{0.01}[/tex]
[tex]z=\frac{0.04}{0.01}[/tex]
[tex]z=4[/tex]
Now we will use formula to find the probability between two z-score as:
[tex]P(a<z<b)=P(z<b)-P(z<a)[/tex]
Upon substituting our given values in above formula we will get,
[tex]P(-4<z<4)=P(z<4)-P(z<-4)[/tex]
Using normal distribution table we will get,
[tex]P(-4<z<4)=0.99997-0.00003[/tex]
[tex]P(-4<z<4)=0.99994[/tex]
Now let us convert our answer into percent by multiplying 0.99994 by 100.
[tex]0.99994\times 100=99.994\%[/tex]
Therefore, 99.994% of screws are between 0.61 and 0.69 inches and option D is the correct choice.
