The weight of turkeys is normally distributed with a mean of 22 pounds and a sta

Question

The weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds. Use Table 1.

a.

Find the probability that a randomly selected turkey weighs between 20 and 26 pounds.(Round "z" value to 2 decimal places and final answer to 4 decimal places.)

b.

Find the probability that a randomly selected turkey weighs below 12 pounds. (Round your answer to 4 decimal places.)

The weight of turkeys is normally distributed with a mean of 22 pounds and a standard deviation of 5 pounds. Use Table 1.

TABLE 1 Standard Normal Curve Areas Entries in this table provide cumulative probabilities, that is, the area under the curve to the left of-z. For example, P(Z-_ 1.52) = 0.0643 0.0001 0.0001 0.0001 0.0001 0.0002 0.0003 0.0001 0.0001 0.0001 0.0002 0.0003 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0003 0.0001 0.0001 0.0001 0.0002 0.0003 0.0002 0.0002 0.0003 0.0005 0.0007 0.0003 0.0005 0.0005 0.0008 0.0005 0.0007 0.0010 0.0010 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 0.0026 0.0035 0.0047 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0025 0.0034 0.0045 0.0023 0.0032 0.0043 0.0023 0.0021 0.0028 0.0038 0.0020 0.0027 0.0041 0.0089 0.0087 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110 0.0179 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183 0.0268 0.0262 0.0244 0.0239 0.0359 0.0436 0.0418 0.0475 0.0582 0.0465 0.0655 0.0630 0.0764 0.0618 0.0778 0.0735 0.0951 0.0901 0.0853 0.0838 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 0.0985 0.1230 0.1210 0.1190 0.1170 0.1587 0.1562 0.1539 01515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1814 0.1788 0.1762 0.1736 0.1711 977 578 3050 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 0.3707 0.3669 0.3632 0.3594 0.3557 0.35 0.3936 0.4602 0.4562 0.4522 0.4483 0.44430.4404 0.43640.4325 0.4286 0.4247 0.4880 SOURE: Probablities calculated with Excel. TABLE 1 (Continued) Entries in this table provide cumulative probabilities, that is, the area under the curve to the left of z. For example, P 1.52) = 0.9357 0.5239 0.5279 0.5319 0.5398 0.5675 0.5714 0.5793 0.6554 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.68440.6879 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.7549 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.8264 0.8365 0.8643 0.8849 0.9032 0.9049 0.9066 0.9082 0.8830 0.9015 0.8729 0.8925 0.9207 0.9222 0.9236 0.9251 0.9265 0.9292 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 0.9452 0.9463 0.94740.9484 0.9495 0.9505 0.9515 0.95250.9535 0.9545 0.9554 0.9564 0.9573 0.9582 0.9591 0.9641 0.9599 0.9633 0.9649 0.9656 0.9664 0.9671 0.9678 0.9719 0.9726 0.9732 0.9738 0.97440.9750 0.9756 0.9761 0.9767 2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 9842 991 993 993 0.9934 0.9938 997 997 0.9984 0.9988 0.9992 0.9986 0.9987 0.9992 0.9993 0.9993 0.9993 0.9995 0.9995 0.9997 0.9997 0.9997 SOURCE: Probablities cakculated with Excel

Explanation / Answer

A.

Convert the weights into z: = (x - mean)/standard dev

For 20 pounds, z = (20-22)/5 = -0.4

For 26 pounds , z = (26 - 22)/5 = 4/5 = 0.8

Probability that the turkey weighs between 20 and 26 pounds = P(z <= 0.8) - P(z <= -.4)

= 0.7881 - 0.3446 = 0.4435

B.Converting to z = (12 - 22)/5 = -2

P (z <= (12-22)/5) = P(z <= -2) = 0.0228

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