On a standardized test with a normal distribution the mean score was 67.2. The standard deviation was 4.6. What percent of the data fell between 62.6 and 71.8?

Question 2 options:

95%

68%

4.6%

13.2%

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%

just put them in order

step-by-step explanation:

no this is a nonlinear negative