Suppose a batch of steel rods produced at a steel plant have a mean length of 178178 millimeters, and a standard deviation of 1212 millimeters. if 7676 rods are sampled at random from the batch, what is the probability that the mean length of the sample rods would differ from the population mean by less than 0.560.56 millimeters? round your answer to four decimal places.

To solve this problem we use the z statistic. The formula for z score is:

z = (x – u) / s

where,

x = sample value mean = 178 – 0.56 = 177.44 mm

u = total sample mean = 178 mm

s = standard deviation = 12 mm

z = (177.44 – 178) / 12

z = -0.05

from the tables P (z = -0.05) = 0.484

So the probability P is: [multiply by 2 since we account the two sides)

P = (0.5 – 0.484) * 2 = 0.0320

b or 2, just plug these equations into your graphing calculator then check the graph to see the image. if it matches up then that's your answer!

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!