Question

Standard deviation 16.11 Mean 51.24 For each of the following values, use the one-tail five percent (5%) criteria and determine if the occurrence of the score would be considered relatively likely or unlikely? (Note: Assume that the researcher is only interested in the extreme low percents of students receiving free or reduced lunch -- or the negative end of the normal curve.)

a. 84.5

b. 21.4

c. 24.1

For each of the following values, use the two-tail five percent (5%) criteria and determine if the occurrence of the score would be considered relatively likely or unlikely?

a. 21.4

b. 84.5

Answer 1

Ans:

1)a)

z=(84.5-51.24)/16.11

z=2.065

P(z>2.065)=0.0195

As,above probability is less than 0.05,so it is unlikely.

b)

z=(21.4-51.24)/16.11

z=-1.852

P(z<-1.852)=0.0320

As,above probability is less than 0.05,so it is unlikely.

c)

z=(24.1-51.24)/16.11

z=-1.685

P(z<-1.685)=0.0460

As,above probability is less than 0.05,so it is unlikely.

2)

a)P(-1.852<z<1.852)=2*P(z<-1.852)=0.0640

As,above probability is greater than 0.05,so it is likely.

b)

P(-2.065<z<2.065)=2*P(z<-2.065)=2*0.0195=0.0390

As,above probability is less than 0.05,so it is unlikely.