Assigment3/Math1131.03/Solution/31
1.a) P(Z 1.76) = 0.9608 [1]
b)
Since the distribution is continuous, P(Z = 1.23) = 0. [1]
c)
P(-1.11 Z 1.35) = P(Z 1.35) – P(Z < -1.11) = P(Z 1.35) – P(Z -1.11) = 0.9115 - 0.1335 = 0.778. [1]
d)P(Z -5.30) is less than P(Z -3.49), which is 0.0002 (to the 4th decimal place). Thus, P(Z -5.30) < 0.0002 [1]
e) P(Z 6.23) is greater than P(Z 3.49), which is .9998, (rounded to the 4th decimal place). Thus P(Z 6.23) > .9998 [1]
2. X (diameter of ball bearing) is a normal random variable with = 2 and = .005.
a) P(X < 2.01) = = P(Z 2) = 0.9772. [1]
b) P(X > 1.99) = = 1 - P(Z -2) = 1 - 0.0228 = 0.9772. [1]
c) P(2.01 < X < 2.013) = = P(2 < Z < 2.6) = P(Z 2.6) – P(Z 2) = 0.9953 – 0.9772 = 0.0181. [1]
d).01 = 2(.005) = 2,so P(X is within 2 of the mean) = P(-2 < Z < 2) = P(Z 2) – P(Z -2) = 0.9772 - 0.0228 = 0.9544. [1]
(1)
3. a) zo = 2.65. [1]
b) Since this is a continuous distribution, 1 – P(Z zo) = 0.40, so P(Z zo) = 0.60. P(Z 0.25) = 0.5987 and P(Z 0.26) = 0.6026. Therefore, zo 0.25. [1]
c) P(-zo Z zo) = 0.60. Since the standard normal curve is symmetric, P(Z -zo) = P(Z zo) = (1 - 0.60)/2 = 0.20. P(Z -0.85) = 0.1977 and P(Z -0.84) = 0.2005, so zo 0.84. [1]
d) P(-2 < Z < zo) = P(Z zo) – P(Z -2) = 0.90 P(Z zo) = 0.90 + 0.0228 = 0.9228. P(Z 1.42) = 0.9222 and P(Z 1.43) = 0.9236, so zo 1.42. [1]
4. X is a normal random variable with = 50 and = 8.
a) P(X xo) = = 0.95. P(Z 1.64) = 0.9495 and P(Z 1.65) = 0.9505. Since these probabilities are equally distant from 0.95, there are three possible answers: 1) zo = = 1.64, so xo = 63.12.
2) zo = = 1.65, so xo = 63.2. [1]
3) zo = (1.64 + 1.65)/2 = 1.645 = , so xo = 63.16.
b) 1 – P(X < xo) = 0.25, so = 0.75. P(Z 0.67) = 0.7486 and P(Z 0.68) = 0.7517, so zo = 0.67 and xo 55.36. [1]
c) P(xo X 64) = – P(X xo) = P(Z 1.75) – P(X xo) = 0.9599 – P(Z zo) = 0.342 P(Z zo) = 0.6179. zo = = 0.30 and xo = 52.4. [1]
d)P(50 – xo X 50 + xo) = 0.85. Since the standard normal curve is symmetric, P(X 50 - xo) = P(X 50 +xo) = (1 - 0.85)/2 = 0.075. P(Z -1.44) = 0.0749 and P(Z -1.43) = 0.0764, so xo 11.52. [1]
(2)
5. X (ACT scores) is a normal random variable with = 20 and = 6.
P(X xo) = 0.05, so P(X < xo) = = 0.95. P(Z 1.64) = 0.9495 and P(Z 1.65) = 0.9505. Since these probabilities are equally distant from 0.95, there are three possible answers: 1) zo = = 1.64, so xo = 29.84.
2) zo = = 1.65, so xo = 29.9. [2]
3) zo = (1.64 + 1.65)/2 = 1.645 = , so xo = 29.87.
A student must attain a score of 30 in order to be in the running for a scholarship.
6.X(number of unfit trucks) is a binomial random variable with n = 500, p = 0.42.
a)i) E(X) = = np = 500(0.42) = 210. [1]
ii) 2 = npq = 500(0.42)(0.58) = 121.8. [1]
= = 11.04.
np = 210 5 and nq = 500(0.58) = 290 5, so we may use a Normal approximation.
b)i) P(X 200) = = P(Z -9.5/11.04) = P(Z -.86) = .1949. [1]
ii) P(X 225) = 1 – P(X 224) = 1 – = 1 – P(Z 14.5/11.04) = 1 – P(Z 1.31) = 1 – 0.9049 = 0.0951. [1]
iii)P(200 X 225) = P(X 225) – P(X 199) = – = P(Z 15.5/11.04) – P(Z -10.5/11.04) = P(Z 1.40) – P(Z -0.95) = 0.9192 – 0.1711 = 0.7481. [1]
(3)
7.a) n = 36 > 30, so we can use the Central Limit Theorem. is approximately normally distributed with = = 65 and = 15/6 = 2.5.
(i) P( 70) = = P(Z 2) = 0.9772. [1]
(ii) P(> 62) = 1 – P( 62) = 1 – = 1 – P(Z -1.2) = 1 – 0.1151 = 0.8849. [1]
(iii) P(67 < < 72) = P(< 72) – P( 67) = – = P(Z 2.8) – P(Z 0.8) = 0.9974 – 0.7881 = 0.2093. [1]
8. Since X is normally distributed, is exactly normally distributed with = = 500 and = 116/5 = 23.2.
(i) P(> 540) = 1 – P( 540) = 1 – = 1 – P(Z 1.72) = 1 – 0.9573 = 0.0427. [1]
(ii) P(< 460) = = P(Z -1.72) = 0.0427. [1]
(iii) P(475 < < 525) = P(< 525) – P( 475) = – = P(Z 1.08) – P(Z -1.08) = 0.8599 – 0.1401 = 0.7598.
[1]
(iv)P( is within 50 of the true mean) = P(450 < < 550) = P(< 550) – P( 450) = – = P(Z 2.16) – P(Z -2.16) = 0.9846 – 0.0154 = 0.9692.[1]
(4)