Respuesta :
Answer:
Mean, [tex]\mu[/tex] = 133.09
Step-by-step explanation:
We are given that the random variable x has a normal distribution with standard deviation 21,i.e;
X ~ N([tex]\mu,\sigma = 21[/tex])
The Z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ (0,1)
Also, it is known that the probability that x exceeds 160 is 0.90 ,i.e;
P(X > 160) = 0.90
P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{160-\mu}{21}[/tex] ) = 0.90
From the z% table we find that at value of x = -1.2816, the value of
P(X > 160) is 90%
which means; [tex]\frac{160-\mu}{21}[/tex] = -1.2816
160 - [tex]\mu[/tex] = 21*(-1.2816)
[tex]\mu[/tex] = 160 - 26.914 = 133.09
Therefore, mean [tex]\mu[/tex] of the probability distribution is 133.09.
