Question

52. Use the following table, Areas in the Tail of the Normal Distribution, to answer the q 0 5000 4960 4920 4880 4840 4801 4761 47214841 1 4602 4562 4522 44834443 4404 4364 4325 4 42 2 |4207 4168 4129 4090 4052 4013 3974: 39% 3esr 3 3821 37833745 3707 36693632 3594 3557 3520 5 3085 3050 3015 2581 2545 212 2877 2843281 6 2743 2709 2676 2643 2611 2578 25462514 7 2420 2389 2358 2327 22962256 2236 3446 3409 3372 3336 3300 IN me 3192 3156 3121 2776 2206 2177 21 2119 2090 2061 2033 2005 1977 1949 1922 18941867 1788 1.0 1587 1562 1539 1515 1492 1469144614231401 1379 1357 1335 1314 1292 1271 1251 1230121011901170 1093 1038 1.3 0968095109340918 0901 0885 0869 0853 0838 0823 1.4 0808 0793 0778 0764 0749 0735 0721 0708 0694 0681 1.5 0668 0655 0643 0630 0618 0606 05940582 0571 0559 1.6 0548 0537 0526 0516 0505 0495 0485 0475 04650455 1.7 0446 0436 0427 0418 0409 0410392 0384 0375 0367 1.8 0359 0351 0344 0336 0329 03220314 0307 0301 02 1.9 0287 02810274 0268 0262 0256 2500244 0239 0233 20 0228 0222 0217 0212 02070202 0197 0192 0188 0183 21 0179 01740170 168 0162 0158 0154 0150 16 0143 22 0139 01360132 0129 0125 0122 0119 0116 0113011 2.3 0107 0104 0102 099 0096 009400910089 0087 0084 24 0082 008070075 0073 0071 0069 0058 0066 0064 2.5 0062 006000570055 0054 0052 0051 004004 26 00470045 004 043 0041 0040 0039 00303 003 27 0035034 03 0032 031 0030029 028 02 00 28 0026 0025 0024 0023 00230022 0021 0021 000 0019 29 0019 0018 0018 0017 0016 0016 0015 0015 014 001 3.0 0013 0013 0013 0012 001200110011 0011010 Last year, Lift and Grunt Gym had an average of 180 client visits per day, with

Answer 1

Probability of the event = P(z)

where z = (x-µ)/σ

given

µ = 180

σ = 25

(A) z = (150 - 180)/25 = -1.2

from the table, for z = -1.2, P = 1 - 0.1151 = 0.8849

Hence, P(z) > 150 = 100 - 88.49 % = 11.51%

(B) z = (100 - 180)/25 = -3.2

from the table, for z = -3.2, P = 1 - 0.0010 = 0.9990

Hence, P(z) < 100 = 100 - 99.99 = 0.01 %

(C) z₁ = (170 - 180)/25 = -0.4

z₂ = (205 - 180)/25 = 1

for z₁ = -0.4, P₁ = 1 - 0.3446 = 0.6554

for z₂ = 1, P₂ = 0.1587

Hence, 170 < P(z) = 65.54%

P(z) > 205 = 100 - 15.87 = 84.13%

Hence, 170 < P(z) < 205 is between 65.54% and 84.13%