Astudent conducted a study and reported that the 95% confidence interval for the mean ranged from 46 to 54. he was sure that the mean of the sample was 50, that the standard deviation of the sample was 16, and that the sample size was at least 30, but could not remember the exact number. can you him out?

Step-by-step explanation:

Given that a student conducted a study and reported that the 95% confidence interval for the mean ranged from 46 to 54

Hence width of interval = 8

Margin of error = ±1/2 (width)=±4

Margin of error = Z critical for 95%( std  error)

Equate the two terms for margin of error we get

(\frac{16}{\sqrt{n}=

n=61.4656

So sample size is approximately 61.

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

20%

step-by-step explanation: