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

D. 99.9936%.

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=(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=(0.61-0.65)/(0.01)`

`z=(-0.04)/(0.01)`

`z=-4`

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

`z=(0.69-0.65)/(0.01)`

`z=(0.04)/(0.01)`

`z=4`

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

`P(a<z<b)=P(z<b)-P(z<a)`

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

`P(-4<z<4)=P(z<4)-P(z<-4)`

Using normal distribution table we will get,

`P(-4<z<4)=0.99997-0.00003`

`P(-4<z<4)=0.99994`

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

`0.99994* 100=99.994\%`

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