Respuesta :
Answer:
Standard deviation of the sample = 17.421
Step-by-step explanation:
We are given that in a random sample of 80 teenagers, the average number of texts handled in one day is 50.
Also, the 96% confidence interval for the mean number of texts handled by teens daily is given as 46 to 54.
So, sample mean, [tex]xbar[/tex] = 50 and Sample size, n = 80
Let sample standard deviation be s.
96% confidence interval for the mean number of texts,[tex]\mu[/tex] is given by ;
96% confidence interval for [tex]\mu[/tex] = [tex]xbar \pm 2.0537*\frac{s}{\sqrt{n} }[/tex]
[46 , 54] = [tex]50 \pm 2.0537*\frac{s}{\sqrt{80} }[/tex]
Since lower bound of confidence interval = 46
So, 46 = [tex]50 - 2.0537*\frac{s}{\sqrt{80} }[/tex]
50 - 46 = [tex]2.0537*\frac{s}{\sqrt{80} }[/tex]
s = [tex]\frac{4*\sqrt{80} }{2.0537}[/tex] = 17.421
Therefore, standard deviation of the sample is 17.421 .
