Question #7d0f8
1 Answer
# mu = 15.4988# (4dp)
# sigma = \ \ 3.0998# (4dp)
Explanation:
:et the mean be
#Y ~ N(mu, sigma^2)#
We are given that the mean is equal to five times the standard deviation. So:
# mu = 5sigma # .... [A]
We also know that:
# P( Y gt 20) = 0.0732 #
From which we can deduce that:
# \ \ \ \ \ P( Z gt (20-mu)/sigma ) = 0.0732 # where#Z ~ N(0, 1)#
# :. P( Z lt (20-mu)/sigma ) = 1-0.0732 = 0.9268 #
Using Normal Distribution Tables, for the function:
# Phi(z) = P(Z lt z) #
we have:
# (20-mu)/sigma = Phi^(-1)(0.9268) #
# :. (20-mu)/sigma = 1.4521 #
# :. 20-mu = 1.4521 sigma # ..... [B]
Eq[A}+Eq[B] gives:
# 20 = 6.4521 sigma => sigma = 3.0998# (4dp)
And using [A] this leads to:
# mu = 15.4988# (4dp)
