Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Could someone do this problem step by step please? Thank you! Students in a mate

ID: 3131645 • Letter: C

Question

Could someone do this problem step by step please? Thank you!

Students in a materials lab are required to experimentally determine the heat conductivity of aluminum. If student-derived values are normally distributed about a mean of .5 (cal/(cm)(sec)(degree C)) with standard deviation of .03, evaluate the probability that an individual student will obtain a conductivity from .48 to .52. If student values have the mean and standard deviation given in (a), evaluate the probability that a class of 25 students will produce a sample mean conductivity from .48 to .52. If student values have the mean and standard deviation given in (a), evaluate the probability that at least 2 of the next 5 values produced by students will be in the range from .48 to .52.

Explanation / Answer

MEAN = 0.5

STANDARD DEV = 0.03

A) P(0.48<X<0.52) =

For x = 0.48 , z = (0.48- 0.5) / 0.03 = -0.66 and for x = 0.52, z = (0.52 - 0.5) / 0.03 = 0.66

Hence P(0.48 < x < 0.52) = P(-0.66 < z < 0.66) = [area to the left of z = 0.66] - [area to the left of -0.66]

= 0.7454 - 0.2546 = 0.4908

B) IN THIS PART AGAIN WE WILL HAVE THE SAME PROBABILITY OF 0.4908

AS THE CLASS SIZE WILL NOT HAVE ANY EFFECT.

C) OUT OF 5 WE NEED AT LEAST TWO

P(ATLEAST 2) = 1- P(0) - P(1)

P(0) = 5C0*(0.49)^0*(0.51)^5 = 0.034

P(1) = 5C1*(0.49)^1*(0.51)^4 = 0.165

TOTAL WILL BE 0.199

HENCE P(AT LEAST 2) = 1 -0.199 = 0.801