Q:

The data to the right represent the weights​ (in grams) of a random sample of 50 candies. Complete parts​ (a) through​ (f). 0.920.92 0.870.87 0.880.88 0.820.82 0.820.82 0.870.87 0.970.97 0.860.86 0.890.89 0.840.84 0.810.81 0.880.88 0.770.77 0.860.86 0.930.93 0.840.84 0.720.72 0.820.82 0.740.74 0.830.83 0.930.93 0.750.75 0.790.79 0.910.91 0.840.84 0.910.91 0.880.88 0.880.88 0.830.83 0.780.78 0.990.99 0.810.81 0.780.78 0.750.75 0.820.82 0.760.76 0.820.82 0.870.87 0.910.91 0.770.77 0.720.72 0.940.94 0.710.71 0.730.73 0.810.81 0.810.81 0.860.86 0.930.93 0.930.93 0.820.82 ​(a) Determine the sample standard deviation weight.

Accepted Solution

A:
Answer:0.069Step-by-step explanation:The given data set is0.92, 0.87, 0.88, 0.82, 0.82, 0.87, 0.97, 0.86, 0.89, 0.84, 0.81, 0.88, 0.77, 0.86, 0.93, 0.84, 0.72, 0.82, 0.74, 0.83, 0.93, 0.75, 0.79, 0.91, 0.84, 0.91, 0.88, 0.88, 0.83, 0.78, 0.99, 0.81, 0.78, 0.75, 0.82, 0.76, 0.82, 0.87, 0.91, 0.77, 0.72, 0.94, 0.71, 0.73, 0.81, 0.81, 0.86, 0.93, 0.93, 0.82.Formula for mean:[tex]Mean=\frac{\sum x}{n}[/tex]Sum of all terms = 41.98Mean of the data set is[tex]Mean=\frac{41.98}{50}[/tex][tex]Mean=0.8396[/tex]Formula for standard deviation for population:[tex]\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}[/tex]Formula for standard deviation for sample:[tex]\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n-1}}[/tex][tex]\sigma=\sqrt{\frac{0.231792}{50-1}}[/tex][tex]\sigma=\sqrt{0.004730449}[/tex][tex]\sigma=0.06877826[/tex][tex]\sigma\approx 0.069[/tex]Therefore, the standard deviation of the data set is 0.069.