Question

Visual sequential memory and poor spellers. Holmes, Malone, and Redenbach (2008) found that good readers and good spellers correctly read 93.8 + 2.9 M SD] words from a spelling list. On the other hand, average readers and poor spellers correctly read 84.8 ± 3.0 (M ±SD] words from the same spelling list. Assuming with a z score proportion of a 32. Can eating slo a meal? In one Privitera, Coop were asked to consumed in t study, particip 627 + 183 kil who ate fast c Assuming thes lal In which these data are normally distributed, lal What percentage of participants correctly read at least 90 words in the good readers and good spellers group? b) What percentage of participants correctly read at least 90 words in the average readers and poor spellers group? group, wa consume

Answer 1

a) Mean = 93.8 and Standard deviation= 2.9 also given that the distribution is normal hence it follows Z statistic , now at 90 words Z score is calculated as

$Z=\frac{X-\bar{X}}{\sigma}=\frac{90-93.8}{2.9}=-1.31$

By .Z statistic table

Percentage that atleast read 90 words is 90.4 9 %

b) Hera also X=90 and also by same formulae

$Z=\frac{X-\bar{X}}{\sigma}=\frac{90-84.8}{3}=-1.733$

By Z table at Z=-1733 the percentage of minimum participants correctly read at least 90 is 95.85

The Z statistic is shown below, here The percentage value that shows in the table is actually the value left to the Z score which is by default , and we need to find the value right to The calculated Z value, hence we subtract the table value with 1 we get the right proportion and then convert into percentage.

Table