In: Statistics and Probability
An instructor is interested in assessing students' typing speed
in the class. He knows that the average at the university is 49
word per minute (wpm) with a SD of 18 wpm. He collects
data from 31 of his students.
a) What is the probability of the average typing
speed being less than 49 wpm?
probability =
b) What is the proability of the average typing
speed being between 51 and 53 wpm?
probability =
Note: Do NOT input probability responses as
percentages; e.g., do NOT input 0.9194 as 91.94.
Let x be the students' typing speed in the class.
We are given mean ( µ ) = 49 and standard deviation (σ) = 18 and n = 31
Therefore according to sampling distribution of sample mean ,
follows
approximately normal distribution with mean (
)=
49
and standard deviation =
=
= 3.2329
= P( z < 0 )
= 0.5000 ------ ( from z score table , value for z= 0 )
Part b) P( 51 ≤ ≤ 53 ) = P(
≤ 53 ) - P(
≤ 51 )
= P( z < 1.24 ) - P( z < 0.62 )
= 0.8925 - 0.7324 ------------ ( from z score table )
= 0.1601