The random variable x has a normal distribution with standard deviation 21. It is known that the probability that x exceeds 160 is .90. Find the mean mu of the probability distribution.

Mean, = 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()

The Z probability is given by;

              Z = ~ (0,1)

Also, it is known that the probability that x exceeds 160 is 0.90 ,i.e;

 P(X > 160) = 0.90

 P( > ) = 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;     = -1.2816

                           160 - = 21*(-1.2816)

                              = 160 - 26.914 = 133.09

Therefore, mean of the probability distribution is 133.09.

i need a picture

step-by-step explanation:

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)