Answer:
Explanation:
Given that the random variable X is normally distributed, with
mean = 50 and standard deviation = 7.
Then we have z=
![(x-50)/(7) is N(0,1)](https://img.qammunity.org/2020/formulas/mathematics/college/jlyrh45ss2ds84jubed6z2reeqlyrdrav9.png)
Using this and normal table we find that
a)
![P(56<x<68)\\= P(0.86<Z<2.57)\\=0.4949-0.3051\\=0.1898](https://img.qammunity.org/2020/formulas/mathematics/college/a8nys7bkpbevj73pwbupb8m27zgv90ro4r.png)
b) When z=0.02
we get
![x=50+0.02(7)=50.14](https://img.qammunity.org/2020/formulas/mathematics/college/qkclzbvj1t9gv3p11v3vix0ao1f36e3tym.png)
c) 90th percentile z value =1.645
90th percentile of X
![=50+7(1.645)\\= 50+11.515\\=61.515](https://img.qammunity.org/2020/formulas/mathematics/college/fxpf5glfnzdnfdabhff4rdd92uzwbm7jss.png)