Respuesta :
Answer:
416
Step-by-step explanation:
Standard deviation is 2.6
Margin error for the problem is 0.25
Probability 95%, that means thet the siginficance level α is 1 – p
α = 1 – 0.95 = 0.05
margin of error (ME) can be defined as follows
ME = Z(α/2) * standard deviation/ √n
Where n is the sample size
Z(0.05/2) = Z(0.025)
Using a z table Z = 1.96
Now, replacing in the equation and find n
.25 = 1.96 * 2.6/ √n
0.25 = 5.096/√n
√n = 5.096 /0.25
√n = 20.38
n = 20.38^2
n = 415.5 = 416
So if standard deviation is around 2.6 Â then 416 students should be sampled to be within 0.25 drinks of population mean with 95% probability.
What is standard deviation?
Standard deviation is a number used to tell how measurements for a group are spread out from the average (mean).
Here given that
standard deviation=2.6
Margin error =0.25
Probability = P=95%=0.95
Now we can calculate level significance level  as
α = 1 – p= 1 – 0.95 = 0.05
Now  sample size can be calculated as
[tex]ME = Z(\alpha/2) \times\frac{ \text{standard deviation}}{\sqrt n}[/tex]
[tex]0.25=Z(0.025)\times\frac{ 2.6}{\sqrt n}[/tex]
Using Z-table
[tex]0.25=1.96\times\frac{ 2.6}{\sqrt n}\\\\\sqrt n= 1.96\times\frac{ 2.6}{0.25}\\\\\\sqrt n=20.38\\\\n=20.38^2\\\\n=415.5[/tex]
But n must be whole number so n=416
So if standard deviation is around 2.6 Â then 416 students should be sampled to be within 0.25 drinks of population mean with 95% probability.
To learn more about standard deviation visit :https://brainly.com/question/12402189
