Question

1. For a distribution, the sample mean is 96, standard deviation is 12, and number of participants is 285. Please show all work.

       

1. What is the relative frequency of scores between 80 and the mean?

2. How many participants are expected to have scores that fall between 80 and the mean?

3. What is the percentile of someone scoring 80?

4. How many participants are expected to have scores that are above 80?

Answer 1

Let X be the distribution of scores.

Given:

For such a large sample size, we can assume its distribution to be a normal distribution.

a. What is the relative frequency of scores between 80 and the mean?

P(80<X<96)

We can standardize this to the standard normal distribution.

P( (80-96)/12 < Z < (96-96)/12 ) = P(-1.33 < Z < 0) = 0.40 [from normal distribution table]

b.How many participants are expected to have scores that fall between 80 and the mean?

We already calculated the frequency of scores between 80 and the mean.

No. of participants = P(80<X<mean)*285 = 0.40*285 = 114

c. What is the percentile of someone scoring 80?

For a score of 80, z value will be

z = (80-96)/12 = -1.33

Till z = -1.33, area under standard normal curve = fraction of data below -1.33 = P(Z<-1.33) = 0.091

Hence perccentile = 0.091*100 = 9.1 percentile

d. How many participants are expected to have scores that are above 80?

Percentile at 80 = 9.1, so above 80, there will be (100-9.1)% of participants = 91.9% of participants

= 0.919*285 = 261.9 = 262 participants