Average talk time between charges of a cell phone is advertised as 4 hours. Assu

Question

Average talk time between charges of a cell phone is advertised as 4 hours. Assume that talk time is normally distributed with a standard deviation of 0.8 hour. Use Table 1.

Find the probability that talk time between charges for a randomly selected cell phone is below 3.5 hours. (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

Find the probability that talk time between charges for a randomly selected cell phone is either more than 4.5 hours or below 3.5 hours. (Round "z" value to 2 decimal places and final answer to 4 decimal places.)

Twenty-five percent of the time, talk time between charges is below the 1st quartile value. What is this value? (Round "z" value to 2 decimal places and final answer to 3 decimal places.)

Average talk time between charges of a cell phone is advertised as 4 hours. Assume that talk time is normally distributed with a standard deviation of 0.8 hour. 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)

If X be the random variable representing mean load weight of sample, then

P(X < 3.5) = P[(X - ) / ) < (3.5 – 4) / 0.8]

Where

: Population mean = 4

: Sample standard deviation = 0.8

So, we have

P(X < 3.5) = P[(X - ) / ) < (3.5 – 4) / 0.8]

= P(Z < - 0.63)

= 0.2643

(b)

P(X > 4.5) = P[(X - ) / ) > (4.5 – 4) / 0.8]

= P(Z > 0.63)

= 1 - P(Z < 0.63)

= 1 - 0.7357

= 0.2643

So, P(Z < 3.5) or P(Z > 4.5) = 0.2643 + 0.2643

= 0.5286

(c)

P = 0.25

When Z = - 0.67, P is closest to 0.25 (= 0.2514).

So required value is Z = - 0.67

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