Answer the following questions.
a. Find the number z such that the proportion of observations that are less than z in a standard Normal distribution is 0.9986. z =
b. Find the number z such that 60% of all observations from a standard Normal distribution are greater than z. z =

Step-by-step explanation:

Hello!

The standard normal distribution is a bell-shaped distribution with mean 0 and standard deviation 1.

a.

You need to find the number of Z so that 0.9986 of the distribution is below it. Symbolically:

P(Z≤a) = 0.9986

To do so you have to use the table of the standard deviation distribution. These tables show values of accumulative probabilities

(right entry, positive values of Z) and

(left entry of the table, negative numbers of Z)Since the probability value greater than 0.5 (remember, for this distribution, the mean is zero and the cumulative probability to the mean is the 50% of the distribution), you have to look for it in the body of the right Z-table and then reach the margins to know the corresponding Z value. The first column shows the integer and first decimal and the first row shows the second decimal.

(1st attachment)

Z= 2.98

b.

Now you have to look for a value of the distribution that has above it 60% of the distribution, symbolically:

P(Z>b)=0.6

Now the table only gives you values "less or equal than" the corresponding value of Z. So you have to rewrite this expression so that you cand find the missing value. A little reminder: The maximal value of probability for any distribution is 1. So you can rewrite the expression as:

P(Z>b)=0.6 ⇒ P(Z≤b)= 1 - 0.6

P(Z≤b)= 0.4

This value of Z accumulates a probability of 40% since it is less than 50% then you have to look for this value in the left entry of the table and b will be a negative value of Z. Just as before, look for the probability in the body of the table and then the corresponding Z value in the margins. (2nd attachment)

Z= - 0.25

I hope this helps!

answer: -2+√x,-2-√x

i hope this , i don't know if you're looking for a pair or not.