24 mi per min = yards per min
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Answer:
The probability that the candidate score is 600 or greater
P(X≥ 600 ) = 0.0228
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Given mean of the Population = 500
Given standard deviation of the Population = 50
Let 'X' be the random variable in normal distribution
Given X = 600
<u><em>Step(ii)</em></u>:-
The probability that the candidate score is 600 or greater
P(X≥ 600 ) = P(z≥2)
= 0.5 - A(2)
= 0.5 -0.4772
= 0.0228
<u><em>Final answer</em></u>:-
The probability that the candidate score is 600 or greater
P(X≥ 600 ) = 0.0228