Question

The data to the right represent the weights (in grams) of a random sample of 50 candies. Complete parts (a) through (f). 0.89 0.85 0.91 0.95 0.75 0.85 0.86 0.75 0.82 0.98 0.86 0.79 0.77 0.75 0.71 0.83 0.82 0.92 0.77 0.73 0.82 0.84 0.82 0.83 0.85 0.87 0.81 0.92 0.77 0.82 0.99 0.73 0.89 0.86 0.87 0.83 0.82 0.85 0.87 0.93 0.82 0.74 0.84 0.89 0.93 0.92 0.84 0.77 0.79 0.83 (a) Determine the sample standard deviation weight gram(s) (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 approximately bell-shaped so the Empirical Rule can be used. O B. The histogram is approximately bell-shaped so the Empirical Rule cannot be used. OC. The histogram is not approximately bell-shaped so the Empirical Rule cannot be used. O D. The histogram is not approximately bell-shaped so the Empirical Rule can be used. Frequency 0 0.7 Weights (9) (c) Use the Empirical Rule to determine the percentage of candies with weights between 0.706 and 0.970 gram. Hint: * = 0.838. ]% (Type an integer or decimal. Do not round.) (d) Determine the actual percentage of candies that weigh between 0.706 and 0.970 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 0.904 gram. % (Type an integer or decimal. Do not round.) (f) Determine the actual percentage of candies that weigh more than 0.904 gram. % (Type an integer or decimal. Do not round.)

Answer 1

n = number of sample = 50

$\bar{x}=\frac{\sum x}{n}=\frac{41.92}{50}=0.8384$

Standard deviation (S)

$S=\sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}$

$S=\sqrt{\frac{\sum (x-0.8384)^2}{50-1}}$

$S=\sqrt{\frac{0.210272}{49}}$

$S=\sqrt{0.004291265}$

$S=0.06550775$

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

$Z_1=\frac{x-\bar{x}}{S}=\frac{0.706 -0.838}{0.066}=-2$

$Z_2=\frac{x-\bar{x}}{S}=\frac{0.970 -0.838}{0.066}=2$

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

$Z_1=\frac{x-\bar{x}}{S}=\frac{0.706 -0.838}{0.066}=-2$

$Z_2=\frac{x-\bar{x}}{S}=\frac{0.970 -0.838}{0.066}=2$

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

$Z_1=\frac{x-\bar{x}}{S}=\frac{0.904 -0.838}{0.066}=1$

according to emiprical rule 16% data lies above 1 standard deviation of the mean .

Answer : 16 %

( 13.5+2.35+0.15 = 16 )

34% 34% 13.5% 13.5% 2.35% 0.15% H-30 2.35%! 0.15% H-20 H-10 U H + 10 + 20 H +30

Part f

Percentage of candies weight more than 0.904

$Z_1=\frac{x-\bar{x}}{S}=\frac{0.904 -0.838}{0.066}=1$

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

#VALUE! X (x-0.8384)^2 0.89 0.002663 0.85 0.000135 0.91 0.005127 0.95 0.012455 0.75 0.007815 0.85 0.000135 0.86 0.000467 0.75 0.007815 0.82 0.000339 0.98 0.020051 0.86 0.000467 0.79 0.002343 0.77 0.004679 0.75 0.007815 0.71 0.016487 0.83 7.06E-05 0.82 0.000339 0.92 0.006659 0.77 0.004679 0.73 0.011751 0.82 0.000339 0.84 2.56E-06 0.82 0.000339 0.83 7.06E-05 0.85 0.000135 0.87 0.000999 0.81 0.000807 0.92 0.006659 0.77 0.004679 0.82 0.000339 0.99 0.022983 0.73 0.011751 0.89 0.002663 0.86 0.000467 0.87 0.000999 0.83 7.06E-05 0.82 0.000339 0.85 0.000135 0.87 0.000999 0.93 0.008391 0.82 0.000339 0.74 0.009683 0.84 2.56E-06 0.89 0.002663 0.93 0.008391 0.92 0.006659 0.84 2.56E-06 0.77 0.004679 0.79 0.002343 0.83 7.06E-05 Total 41.92 0.210272