The length of a screw produced by a machine is normally distributed with a mean of 0.55 inches and a standard deviation of 0.01 inches. what percent of screws are between 0.53 and 0.57 inches?
99.7%
68%
95%
100%

Find the relation between the range, the standard deviation and the media:

0.57 = 0.55 + 0.02 = 0.55 + 2*0.01 = media + 2* standard deviation

0.53 = 0.55 - 0.02 = 0.55 - 2*0.01 = media - 2 * standard deviation

Then, the desired range is media +/- 2 * standard deviation.

In normally distributed functions, this correspond to 95% confidence interval.

Then, the answer is the third option shown, 95%

31 is prime. and it’s prime factorization is just 31.

That’d be 9 years or something