Respuesta :
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex] \bar X =\frac{58+62+66+70+74+78+54}{7}=66[/tex]
[tex]\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n}}[/tex]
[tex] \sigma =\sqrt{\frac{(58-66)^2 +(62-66)^2 +(66-66)^2 +(70-66)^2 +(74-66)^2+ (78-66)^2 + (54-66)^2}{7}}= 8[/tex]
b. 66,8
Step-by-step explanation:
We have the following data given:
58 , 62 ,66 ,70 , 74 , 78 , 54
Since the data can be modelled with a normal distribution then the best estimator for the true mean is the sample mean given by:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex] \bar X =\frac{58+62+66+70+74+78+54}{7}=66[/tex]
And we can estimate the population deviation with the following formula:
[tex]\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n}}[/tex]
And replacing we got:
[tex] \sigma =\sqrt{\frac{(58-66)^2 +(62-66)^2 +(66-66)^2 +(70-66)^2 +(74-66)^2+ (78-66)^2 + (54-66)^2}{7}}= 8[/tex]
And the best solution for this case is:
b. 66,8
