SOLUTION :
Let the value of z be a so that P(z > a) = 73% = 0.73
Value of a will be negative as it lies to the left of z = 0 and P(z > 0) = 50%.
So,
P(z < a) = 1 - 0.73 = 0.27
From cumulative ND table :
a = - 0.613
Hence, z = - 0.613 to the right of which 73% standard normal curve lies. (ANSWER)
Graph :
2nd graph on the top row represents the given case. Where z = - 0.613 and area to the right of z is 73%.
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