Question

suppose a normal distribution has a mean of 98 and standard deviation of 6. what is p(x ≤ 104)?

Answer 1

Answer:

P(x ≤ 104) = 0.84134

Step-by-step explanation:

We are given that a normal distribution has a mean of 98 and standard deviation of 6 i.e., $\mu$ = 98  and  $\sigma$ = 6

The Z score transformation of normal distribution is ;

           Z  =  $\frac{x-\mu}{\sigma}$ ~ N(0,1)

So, P(x <= 104) = P( $\frac{x-\mu}{\sigma}$ <= $\frac{104-98}{6}$ ) = P(Z <= 1) = 0.84134

From Z table we have calculate the above probability.