Question

The 90th percentile of a normal distribution is how many standard deviations above the mean?

Answer 1

z=1.28
the 90th percentile of a normal distribution is 1.28

Answer 2

The  90th percentile of a normal distribution is 1.28.

We know that

• $\mu = 0$ (Population mean).
• $\sigma = 1$ (Population standard deviation).
• $P(z<z_0) = 0.9000$.
• $P(z<z_0)$ = Cumulative probability value to the left.

Now we determine the value of $z_0$ via using the cumulative standard normal distribution table.

Based on the above table

= 1.2 + 0.08

= 1.28

Therefore we can conclude that the  90th percentile of a normal distribution is 1.28.

Learn more: brainly.com/question/8696744