Respuesta :
Answer:
[tex]\bar X =\frac{2.3+3.1+2.8+1.7+0.9+4+2.1+1.2+3.6+0.2+2.4+3.2}{12}=2.29[/tex]
[tex] s=1.09[/tex]
Step-by-step explanation:
For this case we have the following data given:
2.3 3.1 2.8
1.7 0.9 4.0
2.1 1.2 3.6
0.2 2.4 3.2
Since the data are assumedn normally distributed we can find the standard deviation with the following formula:
[tex]\sigma =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}[/tex]
And we need to find the mean first with the following formula:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex]\bar X =\frac{2.3+3.1+2.8+1.7+0.9+4+2.1+1.2+3.6+0.2+2.4+3.2}{12}=2.29[/tex]
And then we can calculate the deviation and we got:
[tex] s=1.09[/tex]
Answer:
X ~ Norm ( 2.29167 , 1.09045^2 )
Step-by-step explanation:
Solution:-
- The (GPA) for 12 randomly selected college students are given as follows:
2.3 , 3.1 , 2.8 , 1.7 , 0.9 , 4.0 , 2.1 , 1.2 , 3.6 , 0.2 , 2.4 , 3.2
- We are to assume the ( GPA ) for the college students are normally distributed.
- Denote a random variable X: The GPA secured by the college student.
- The normal distribution is categorized by two parameters:
- The mean ( u ) - the average GPA of the sample of n = 12. Also called the central tendency:
[tex]Mean ( u ) = \frac{\sum _{i=1}^{\ 12 }\: Xi }{n}[/tex]
Where,
Xi : The GPA of the ith student from the sample
n: The sample size = 12
[tex]Mean ( u ) = \frac{2.3 + 3.1 + 2.8 + 1.7 + 0.9 + 4.0 + 2.1 + 1.2 + 3.6 + 0.2 + 2.4 + 3.2}{12} \\\\Mean ( u ) = \frac{27.5}{12} \\\\Mean ( u ) = 2.29167[/tex]
- The other parameter denotes the variability of GPA secured by the students about the mean value ( u ) - called standard deviation ( s ):
[tex]s = \sqrt{\frac{\sum _{i=1}^{\ 12 }\: [ Xi - u]^2}{n} } \\\\\\\sum _{i=1}^{\ 12 }\: [ Xi - u]^2 = ( 2.3 - 2.29167)^2 + ( 3.1 - 2.29167)^2 + ( 2.8 - 2.29167)^2 + ( 1.7\\\\ - 2.29167)^2+ ( 0.9 - 2.29167)^2 + ( 4 - 2.29167)^2 + ( 2.1 - 2.29167)^2 + ( 1.2 - 2.29167)^2 +\\\\ ( 3.6 - 2.29167)^2 + ( 0.2 - 2.29167)^2 + ( 2.4 - 2.29167)^2 + ( 3.2 - 2.29167)^2 \\\\\\\sum _{i=1}^{\ 12 }\: [ Xi - u]^2 = 14.26916 \\\\\\s = \sqrt{\frac{ 14.26916 }{12} } \\\\s = \sqrt{1.18909 } \\\\s = 1.09045[/tex]
- The normal distribution for random variable X can be written as:
X ~ Norm ( 2.29167 , 1.09045^2 )
