Question #218373
a) The weekly wages of employees of Volta gold are normally distributed about a mean of $1,250 with a standard deviation of $120. Find the probability of an employee having a weekly wage lying;
i) Between $1,320 and $970
ii) Under $1,400
iii) Over $1,290

b) An achievement test was administered to a class of 20,000 students. The mean score was 80 and the standard deviation was 11. If Lingard scored 72 in the test, how many students did better than him?
1
Expert's answer
2021-07-19T14:13:02-0400

a)

μ=1250σ=120i.  P(970<X<1320)=P(X<1320)P(X<970)=P(Z<13201250120)P(Z<9701250120)=P(Z<0.5833)P(Z<2.333)=0.720050.00982=0.71023ii.  P(X<1400)=P(Z<14001250120)=P(Z<1.25)=0.8943iii.  P(X>1290)=1P(X<1290)=1P(Z<12901250120)=1P(Z<0.3333)=10.63043=0.36957\mu=1250 \\ \sigma=120 \\ i. \; P(970<X<1320) = P(X<1320) -P(X<970) \\ = P(Z< \frac{1320-1250}{120}) -P(Z< \frac{970-1250}{120}) \\ = P(Z< 0.5833) -P(Z< -2.333) \\ = 0.72005-0.00982 \\ = 0.71023 \\ ii. \;P(X<1400) = P(Z< \frac{1400-1250}{120}) \\ =P(Z< 1.25) \\ = 0.8943 \\ iii. \; P(X>1290) = 1 -P(X<1290) \\ = 1 -P(Z< \frac{1290-1250}{120}) \\ = 1 -P(Z<0.3333) \\ = 1 -0.63043 \\ = 0.36957

b)

N=20000μ=80σ=11P(X>72)=1P(X<72)=1P(Z<728011)=1P(Z<0.7272)=10.2336=0.7664n(X>72)=20000×0.7664=15328N=20000 \\ \mu=80 \\ \sigma=11 \\ P(X>72) = 1 -P(X<72) \\ = 1 -P(Z< \frac{72-80}{11}) \\ = 1 -P(Z < -0.7272) \\ = 1 -0.2336 \\ = 0.7664 \\ n(X>72) = 20000 \times 0.7664 = 15328

Answer: 15328 students were better than Lingard


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