Mathematics: Whoever answers first will get brainliest

Whoever answers first will get brainliest

Ответ:

caylah6101

04.09.2021

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.

$z=\frac{x-\mu}{\sigma}$ , where,

$z=\text{z-score}$ ,

$x=\text{Raw-score}$ ,

$\mu=\text{Mean}$ ,

$\sigma=\text{Standard deviation}$ .

Now let us find z-score corresponding to our given raw scores.

$z=\frac{0.61-0.65}{0.01}$

$z=\frac{-0.04}{0.01}$

z=-4

Now let us find z-score for raw score 0.69.

$z=\frac{0.69-0.65}{0.01}$

$z=\frac{0.04}{0.01}$

z=4

Now we will use formula to find the probability between two z-score as:

P(a

Upon substituting our given values in above formula we will get,

P(-4

Using normal distribution table we will get,

P(-4

Now let us convert our answer into percent by multiplying 0.99994 by 100.

$0.99994\times 100=99.994\%$

Therefore, 99.994% of screws are between 0.61 and 0.69 inches and option D is the correct choice.

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