Question

Suppose that z-scores come from a standard normal distribution. Use the Normal Probability Applet (Links to an external site.) to answer this question. What proportion of the z-scores will be between –1 and 1? (Give your answer as a decimal, accurate to three decimal places e.g. If the answer was 0.389184 you would enter 0.389)

Answer 1

Answer:

0.683

Step-by-step explanation:

We have to find P(-1<z<1).

For this purpose, we use normal distribution area table

P(-1<z<1)=P(-1<z<0)+(0<z<1)

Using normal area table and looking the value corresponds to 1.0, we get

P(-1<z<1)=0.3413+0.3413

P(-1<z<1)=0.6826

Rounding the answer to three decimal places

P(-1<z<1)=0.683

So, 68.3% of the z-scores will be between -1 an 1.