The prime factorization of 56
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Answer:
The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.
Step-by-step explanation:
Use the trick of transforming the given random variable one that is standard normal distributed, aka a z-score, then look up the two percentile values in z-score tables to get two equations with two unknowns.
So, let x be the variable describing the marks of a student, and
the standardized equivalent of that (mu - mean, sigma - standard deviation).
We are looking for values of mu and sigma. At this point we'd be out of luck, but, wait, we're given two bits of info: the 10% point (aka, 90th percentile) and the 20% point (can be interpreted as 100-20th percentile). For each point we can use z-score tables to look up the corresponding values of z (just search for z tables). I found:
z-score for the 10% point: z_10=1.28
z-score for the 20% point: z_20=-0.84
That gives us two equations:
and can be solved for mu and sigma (do the work on your end, I am showing my result):
The mean of the distribution is about 53.9 and standard deviation is about 16.5 marks.