I really need help. Please! In a SRS of 80 teenagers, the average number of text messages received per day is found to 50. If the standard deviation of number of texts received is known to be σ = 12, find the 95% confidence interval for the number of text messages received
|dw:1464179953348:dw| Find out the value of +/- Z that enclose 95% of the area under the normal distribution curve. using \(Z=(x-\mu)/\sigma\), => \(x=\sigma Z+\mu\) find the upper and lower limits of x, corresponding to +Z and -Z respectively.
@mathmate Just to be sure are my numbers right to solve the problem? (80-50)/12=2.5 12(1.96)+80=103.52
You need to first find Z that encompasses 95% of the area under the normal distribution curve. Use help from the following link, if necessary. https://www.mathsisfun.com/data/standard-normal-distribution.html
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