I am currently stuck on two questions that I am doing to review for an in class test. Old tests were provided, along with the answer, but lacking the work. If anyone could please explain how to get started I would really appreciate it!
4. A machine cuts circular filters from large rolls of material. If 7.3% of the filters fail to meet specifications, use the normal approximation to the binomial to compute the probability that a sample of 100 of the filters will contain 5 or fewer that fail to meet specifications. ANSWER: 0.2451
6. A test-scoring machine is known to record an incorrect grade on 5% of the exams it grades. Find, by the appropriate method, the probability that the machine records
a. 2 wrong grades in a set of 10 exams.
b. no more than 2 wrong grades in a set of 10 exams.
c. no more than 2 wrong grades in a set of 15 exams.
d. no more than 2 wrong grades in a set of 150 exams.
ANSWER:
a. P(2 wrong in 10) = P[x = 2 | B(n = 10, p = 0.05)] = 0.075
b. P(no more than 2 wrong in 10) = P[x = 0, 1, 2 | B(n = 10, p = 0.05)] = 0.599 + 0.315 + 0.075 = 0.989
c. P(no more than 2 wrong in 15) = P[x = 0,1, 2 | B (n = 15, p = 0.05)] = 0.463 + 0.366 + 0.135 = 0.964
d. P(no more than 2 wrong in 150) = P[x £ 2 | B(150,0.05)] mean= np (150)(0.05) = 7.5 standard deviation = npq = (150)(0.05)(0.95) = 2.6693 Then, P(x less than or equal to 2) = P(x < 2.5) = P[z < (2.5 – 7.5)/2.6693]= P[z < -1.87]= 0.0307
I am aware that the numbers are given (ie 0.599 for b) but I do not know where they got them from. I understand d (a bit) so that one I don't really need explaining.
4. A machine cuts circular filters from large rolls of material. If 7.3% of the filters fail to meet specifications, use the normal approximation to the binomial to compute the probability that a sample of 100 of the filters will contain 5 or fewer that fail to meet specifications. ANSWER: 0.2451
6. A test-scoring machine is known to record an incorrect grade on 5% of the exams it grades. Find, by the appropriate method, the probability that the machine records
a. 2 wrong grades in a set of 10 exams.
b. no more than 2 wrong grades in a set of 10 exams.
c. no more than 2 wrong grades in a set of 15 exams.
d. no more than 2 wrong grades in a set of 150 exams.
ANSWER:
a. P(2 wrong in 10) = P[x = 2 | B(n = 10, p = 0.05)] = 0.075
b. P(no more than 2 wrong in 10) = P[x = 0, 1, 2 | B(n = 10, p = 0.05)] = 0.599 + 0.315 + 0.075 = 0.989
c. P(no more than 2 wrong in 15) = P[x = 0,1, 2 | B (n = 15, p = 0.05)] = 0.463 + 0.366 + 0.135 = 0.964
d. P(no more than 2 wrong in 150) = P[x £ 2 | B(150,0.05)] mean= np (150)(0.05) = 7.5 standard deviation = npq = (150)(0.05)(0.95) = 2.6693 Then, P(x less than or equal to 2) = P(x < 2.5) = P[z < (2.5 – 7.5)/2.6693]= P[z < -1.87]= 0.0307
I am aware that the numbers are given (ie 0.599 for b) but I do not know where they got them from. I understand d (a bit) so that one I don't really need explaining.