Answer:
68.26%
Step-by-step explanation:
This problem can be solved by two methods by utilizing empirical rule and by using normal distribution table. The empirical rule states that
1.The area within one standard deviation contains 68.26% of the data
2.The area within two standard deviation contains 95.44% of the data
3.The area within three standard deviation contains 99.73% of the data
Here μ=76.4 and σ=6.1
μ-σ=76.4-6.1=70.3
μ+σ=76.4+6.1=82.5
The area within one standard deviation contains 68.26% of the data and so the 70.3 points and 82.5 contains 68.26% of the data.
This problem can be solved by using normal distribution table
P(70.3<X<82.5)=?
P(70.3<X<82.5)=P(70.3-76.4/6.1<Z<82.5-76.4/6.1)=P(-1<z<1)=P(-1<z<0)+P(0<z<1)
Using normal distribution we get
P(70.3<X<82.5)=0.3413+0.3413=0.6826
The percentage of the data is 0.6826*100=68.26%
The percentage of the data is between 70.3 points and 82.5 points is 68.26%
z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\x=raw\ score, \mu=mean,\sigma=standard \ deviation[/tex]
[tex]z=\frac{70.3-76.4}{6.1}=-1[/tex]
[tex]z=\frac{82.5-76.4}{6.1}=1[/tex]
P(70.3 < x < 82.5) = P(-1 < z < 1) = P(z < 1) - P(z < -1) = 0.8413 - 0.1587 = 0.6826 = 68.26%
The percentage of the data is between 70.3 points and 82.5 points is 68.26%
Find out more on z score at: https://brainly.com/question/25638875
