35+35+35+35+35+35+35+35+35+35+35+35
Or
35×12
=420
Answer:
P ( -1 < Z < 1 ) = 68%
Step-by-step explanation:
Given:-
- The given parameters for standardized test scores that follows normal distribution have mean (u) and standard deviation (s.d) :
u = 67.2
s.d = 4.6
- The random variable (X) that denotes standardized test scores following normal distribution:
X~ N ( 67.2 , 4.6^2 )
Find:-
What percent of the data fell between 62.6 and 71.8?
Solution:-
- We will first compute the Z-value for the given points 62.6 and 71.8:
P ( 62.6 < X < 71.8 )
P ( (62.6 - 67.2) / 4.6 < Z < (71.8 - 67.2) / 4.6 )
P ( -1 < Z < 1 )
- Using the The Empirical Rule or 68-95-99.7%. We need to find the percent of data that lies within 1 standard about mean value:
P ( -1 < Z < 1 ) = 68%
P ( -2 < Z < 2 ) = 95%
P ( -3 < Z < 3 ) = 99.7%
