Question

The distribution of the number of annual sick days students is normal in company A. An employee named John took 21 number of sick days. The mean number of sick days in company A is 24 and the standard deviation is 1.3. How many fewer days of sick days did John take in standard deviation units? What percentage of employees took fewer sick days that John did?

Answer 1

the distribution of the number of annual sick days is normal with mean 24 days and standard deviation 1.3

let X denotes the number of annual sick days in company A

then X~N(24,1.3²)

an employee named John took 21 number of sick days.

so John took 24-21=3 fewer days of sick days.

so in standard deviation units he took 3/1.3=2.30769 fewer sick days.

the probability of taking fewer sick days than John is

P[X<21]=P[(X-24)/1.3<(21-24)/1.3]=P[Z<-2.30769]=0.01050819

hence 0.01050819*100=1.050819% of employees took fewer sick days that John did.