$63,000 is what percent of $90,000?
2 answers:
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Answer:
Option D
Step-by-step explanation:
Given that a producer ships boxes of produce to individual customers.(say x)
X is a normal random variable with mean = 36 and std dev =4 lbs
By definition of std normal variate we know that
is N(0,1)
75th percentile of std normal distribution is
0.675
Hence corresponding x would be
Option D matches with this value
Hence answer is option D