To solve, you first want to find the total she hiked in the morning. 1.75+2.4=4.15. Then, you want to subtact the total miles she hiked in the morning from the total miles she hiked in total. 6.5-4.15=2.35. The last step is to compare the miles hiked in the morning compared to the miles in the afternoon by finding the difference. 4.15-2.35=1.8, so therefore she hiked 1.8 more miles in the morning than in the afternoon,
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%
That is 2(1/7 + 1/2)
= 2( 2/14 + 7/14)
= 2 * 9/14
= 18/14 = 9/7 units ( or 1 2/7 units)
