The random variable x has a normal distribution with standard deviation. It is known that the probability that x exceeds is. 90. Therefore, mean of the probability distribution is 133.09.
We are given that the random variable x has a normal distribution with standard deviation 21,i.e;
X ~ N( u, σ=21)
The Z probability is given by;
Z = X- u/σ ~ (0,1)
Also, it is known that the probability that x exceeds 160 is 0.90 ,i.e;
P(X > 160) = 0.90
P( X-u/σ > 160-u /21 ) = 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; 160-u /21 = -1.2816
160 - = 21*(-1.2816)
= 160 - 26.914 = 133.09
Therefore, mean of the probability distribution is 133.09.
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