n = number of sample = 50
Standard deviation (S)
Standard deviation
S = 0.066
Part b .
A.The histogram is approximately bell shaped so empirical rule can be used.
Part C
Percentage of candies weight between 0.706 to 0.970
according to emiprical rule 95% data lies between 2 standard deviation of the mean .
Answer : 95 %
Part d
Percentage of candies weight between 0.706 to 0.970
To find actual percentage use Z table or excel command =NORMSDIST(Z score) .
P( Z < -2) = 0.02275
P( Z< 2) = 0.9772499
P( -2 < Z < 2) = 0.9772499-0.02275 = 0.9544997
Answer : 95.450 %
Part E
Percentage of candies weight more than 0.904
according to emiprical rule 16% data lies above 1 standard deviation of the mean .
Answer : 16 %
( 13.5+2.35+0.15 = 16 )
Part f
Percentage of candies weight more than 0.904
To find actual percentage use Z table or excel command = 1-NORMSDIST(1) .
P( Z > 1) = 1- 0.8413447
Answer : 15.866 %
Standard deviation calculation
The data to the right represent the weights (in grams) of a random sample of 50...
The data to the right represent the weights (in grams) of a random sample of 50 candies. 0.82 0.83 0.91 0.99 0.71 0.87 0.88 0.76 0.85 0.98 0.87 0.79 0.78 0.77 0.73 0.85 0.89 0.94 0.78 0.75 0.82 0.98 0.81 0.83 0.81 0.89 0.81 0.91 0.78 0.81 0.98 0.72 0.88 0.82 0.89 0.84 0.84 0.89 0.88 0.94 0.81 0.71 0.84 0.85 0.94 0.85 0.82 0.77 0.76 0.84 Complete parts (a) through (f). (a) Determine the sample standard deviation weight. grams)...
The following data represent the weights (in grams) of a random sample of 50 candies. 0.82 0.81 0.91 0.95 0.74 0.89 0.87 0.75 0.81 0.95 0.88 0.78 0.76 0.75 0.71 0.84 0.87 0.91 0.75 0.75 0.85 0.96 0.85 0.81 0.83 0.88 0.83 0.94 0.76 0.83 0.97 0.71 0.85 0.81 0.85 0.81 0.83 0.86 0.86 0.91 0.82 0.71 0.83 0.86 0.92 0.82 0.82 0.78 0.77 0.81 (a) Determine the sample standard deviation weight. s= 0.07 gram (Round to two decimal places...
I need help with question (d) T 0.82 The following data represent the weights (in grams) of a random sample of 50 candies. 0.89 0.87 0.87 0.84 0.95 0.81 0.94 0.97 0.75 0.76 0.75 0.78 0.84 0.85 0.82 0.92 0.78 0.72 0.83 0.81 0.82 0.82 0.82 0.87 0.81 0.91 0.78 0.81 0.96 0.72 0.86 0.84 0.86 0.94 0.74 0.82 0.87 0.92 0.89 0.86 0.91 0.85 0.84 0.79 0.77 0.81 0.78 0.87 0.74 (a) Determine the sample standard deviation weight. S=...
The following data represent the weights (in grams) of a random sample of 50 candies. 0.82 0.81 0.91 0.95 0.74 0.89 0.87 0.75 0.81 0.95 0.88 0.78 0.76 0.75 0.71 0.84 0.87 0.91 0.75 0.75 0.85 0.96 0.85 0.81 0.83 0.88 0.83 0.94 0.76 0.83 0.97 0.71 0.85 0.81 0.85 0.81 0.83 0.86 0.86 0.91 0.82 0.71 0.83 0.86 0.92 0.82 0.82 0.78 0.77 0.81 (a) Determine the sample standard deviation weight. S= gram (Round to two decimal places as...
I need help at the bottom with - (d) Determine the actual percentage of candies that weigh between 0.7 and 0.98 gram, inclusive - Thank you! The following data represent the weights (in grams) of a random sample of 50 candies. 0.85 0.89 0.82 0.81 0.85 0.87 0.96 0.89 0.91 0.91 0.81 0.86 0.77 0.89 0.85 0.84 0.75 0.84 0.71 0.84 0.91 0.76 0.77 0.95 0.83 0.91 0.88 0.85 0.83 0.78 0.97 0.84 0.75 0.75 0.81 0.76 0.86 0.87 0.86...
I need help with the final part - (f) Determine the actual percentage of candies that weigh more than 0.91 gram. - Thank you! The following data represent the weights (in grams) of a random sample of 50 candies. 0.85 0.89 0.82 0.81 0.85 0.87 0.96 0.89 0.91 0.91 0.81 0.86 0.77 0.89 0.85 0.84 0.75 0.84 0.71 0.84 0.91 0.76 0.77 0.95 0.83 0.91 0.88 0.85 0.83 0.78 0.97 0.84 0.75 0.75 0.81 0.76 0.86 0.87 0.86 0.79 0.73...
Please answer each of the parts in the following question. Show all steps. Thanks in advance! The following data represent the weights (in grams) of a random sample of 50 candies. 0.86 0.84 0.93 0.99 0.73 0.85 0.87 0.77 0.82 0.83 0.82 0.78 0.76 10.75 0.75 0.84 0.86 0.93 0.77 0.74 0.83 0.99 0.83 0.83 0.85 0.86 0.81 0.95 0.79 0.83 0.96 0.75 0.88 0.86 0.88 0.85 0.82 0.87 0.86 0.91 0.87 0.74 0.83 0.85 0.93 0.87 0.84 0.74 0.77...
Question Help The following data represent the weights (in grams) of a random sample of 50 candies. 0.85 0.89 0.82 0.81 0.85 0.87 0.96 0.89 0.91 0.91 0.81 0.86 0.77 0.89 0.85 0.84 0.75 0.84 0.71 0.84 0.91 0.76 0.77 0.95 0.83 0.91 0.88 0.85 0.83 0.78 0.97 0.84 0.75 0.75 0.81 0.76 0.86 0.87 0.86 0.79 0.73 0.95 0.73 0.71 0.83 0.82 0.88 0.93 0.91 0.83 (a) Determine the sample standard deviation weight. = __?__ gram (Round to two...
i need help with with c and d. im using the T1-84 calculator for my class. thank you X23. The Empirical Rule The following data represent the weights (in grams) of a random sample of 50 M&M plain candies. 0.87 0.88 0.82 0.90 0.90 0.84 0.84 0.91 0.94 0.86 0.86 0.86 0.88 0.87 0.89 0.91 0.86 0.87 0.93 0.88 0.83 0.95 0.87 0.93 0.91 0.85 0.91 0.91 0.86 0.89 0.87 0.84 0.88 0.88 0.89 0.79 0.82 0.83 0.90 0.88 0.84...
0.90 0.81 0.88 0.82 0.90 0.84 0.84 0.91 0.94 0.86 0.86 0.86 0.88 0.87 0.89 0.91 0.86 0.87 0.93 0.88 0.83 0.95 0.87 0.93 0.91 0.85 0.89 0.91 0.91 0.86 0.87 0.84 0.78 0.88 0.88 0.89 0.82 0.83 0.8 0.90 0.88 0.84 0.93 0.90 0.86 0.88 0.92 0.85 0.84 0.84 Another measure of central tendency is the trimmed mean. It is computed by determining the mean of a data set after deleting the smallest and largest observed values. Compute the...