Question

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. What percent of screws are between 0.61 and 0.69 inches?
HELP

A. 65%
B. 68%
C. 99.7%
D. 99.9936%

Answer 1

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%

Answer 2

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.

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

$P(-4$

$P(-4$

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.