Scores on an English test are normally distributed with a mean of 36.8 and a standard deviation of 7.6 Find the score that separates the top 59% from the bottom 41% please help ASAP

Probability Density function for normal distribution is given as

F(x) * sigma * sqrt(2*pi) = exp( -(x-mu)^2 / (2* sigma^2) )

log {1 / ( F(x) * sigma * sqrt(2*pi) )} = -(x-mu)^2 / (2* sigma^2)

(2* sigma^2) * log{1 / ( F(x) * sigma * sqrt(2*pi) )} = -(x-mu)^2

sqrt ( - (2* sigma^2) * log{1 / ( F(x) * sigma * sqrt(2*pi) )} )+ mu = x

Finding the score means we need to find the value of x:

x = sqrt ( - (2*7.6^2)* log [1/ (0.41 * 7.6 * sqrt(2*pi)) ] + 36.8

Solving we get value of x to be equal to 35.07

The score that separates top 59% is 35.07

you can also try =normsinv(0.41,36.8,7.6) in excel and get the answer straight away....

Thank you for the help

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