Question

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) (Do not round until the final answer. Then round to three decimal places as needed.) (b) On the basis of the histogram to the right, comment on the appropriateness of using the Empirical Rule to make any general statements about the weights of the candies. O A. The histogram is not approximately bell-shaped so the Empirical Rule can be used. OB. The histogram is approximately bell-shaped so the Empirical Rule can be used. O C. The histogram is not approximately bell-shaped so the Empirical Rule cannot be used. OD. The histogram is approximately bell-shaped so the Empirical Rule cannot be used. CIE 07 Weights (gi (c) Use the Empirical Rule to determine the percentage of candies with weights between 0.699 and 0.983 gram. Hint: x=0.841. % (Type an integer or decimal. Do not round.) (d) Determine the actual percentage of candies that weigh between 0.699 and 0.983 gram, inclusive. % (Type an integer or decimal. Do not round.) (e) Use the Empirical Rule to determine the percentage of candies with weights more than 0.912 gram. % (Type an integer or decimal. Do not round.) If Datermine the actual DACANACA of candies that waich more than 1912 am

Answer 1

a)

∑x = 42.07

n = 50

∑(x-x̅)² = 0.248602

Mean, x̅ = Ʃx/n = 42.07/50 = 0.841

Standard deviation, s = √(Ʃ(x-x̅)²/(n-1)) = √(0.2486/(50-1)) = 0.071

b)

The histogram is approximately bell shaped so the empirical rule can be used.

c)

z scores are:

z = (X-µ)/σ = (0.699-0.841)/0.071 = -2

z = (X-µ)/σ = (0.983-0.841)/0.071 = 2

The empirical rule states that for a normal distribution 95% of the data will fall within two standard deviations of the mean.

Answer: 95%

d)

Actual percentage = 49/50 = 98%

e)

z = (X-µ)/σ = (0.912-0.841)/0.071 = 1

Percentage = 50- 68/2 = 16%