Question

PSYC 1303 Spring 2020: Exam 1 1. The variance and the standard deviation are both measures of variability. Although they are related, there is a key difference between them. Write the equation for both and identify this key difference. Also explain the impact it has on how you interpret each result. 2. Use the following data to calculate the mean and standard deviation. XI ? 3. Assume you have data that is normally distributed with a mean of 555 and a standard deviation of 134. Convert the following raw scores into z scores: 407 750. What proportion of scores fall between those two points? SEIKI

Answer 1

Ans The varience and standard deviation are used to describe the measure of spread around a data set Equations. IF M = mean x = Each value in the o na number of values then data set. in the data set :) varience = f(x-M)2 i Standard deviation = (x;-M)2 varience. * key differences Varience standard deviation The unit of varience is not as same as whit of data The unit of standard deviation is same as the unit of data. of It is the average Squared deviation It is the root mean Square deviation It indicates how far lindividuals in a group Tas Scannedewithpread out CamScanner It indicates observ- ations of a data set differ from its mean.

Interpretation of varience and standard deviation. A large varience has more spread Ivarience indicate Jess spread indicates the data where a small that data bas. A large standard deviation indicates that the data values are far from the mean and a small standard deviation indicates that they are cluster closely around the mean. Here given data is Mean = ² Xi = 6te-3 - 5 3 = 1.6667 standard deviatio :-M12 e = 0.4445+.0.111 + 21,778 = 7.4445 = 2.7284 deviation = 2.7284 unit. Standard CamScanner scanned with

(③ Here xun(41=555 62 (1342) convert following raw score into Now to Z- Scores 2 = x-ul UN(0,1) X=407 Zi = 407-555 - 134 7 = -1. 1045 X = 750 T2 = 750-555 134 72 = 1.4552/ proportion of score fall between two points. P[ 407 <x< 750] =P [-1.10 45 LZ < 1.4552] = [ <1.45] ap[2R-1.04 ] - 1-P [z.> 1.45] - P[2o1.04] - - 0.07352 - 0.13468 -1.104 0 1045 Normal is symmetric, PEZ > 1004] = P [2<-104] Hence P[407<X17507= 10.2082 = 0.7925 Hence proportion of Ses Scantedt withd 750 is CamScanner score 0.7925 fall between