Question

e. 94.52% A bag of chocolate balls contains a mean of 200 balls with a standard deviation of 5 balls. It is normally distributed. The percentage of packets that contains not more than 196 balls would be: Select one: a. 78.81% O b. 0.8% c. 28.81% d. None of the answers offered. O e. 21.19%

Answer 1

The correct answer is option e

To calculate the probability of the percentage of packets not having more than 196 balls, we have to find the Z score

Z=- X - mean SD

196 - 200 Z=

Z = -0.8

By looking at the normal distribution table, the value of probability for a Z score of -0.8 = 0.21186

Therefore the percentage of packets that contains not more than 196 balls would be 21.19%

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The correct answer is option a

Coefficient of variation = Standard deviation Mean

The expected sales = 0.30 $\times$ 10 + 0.40 $\times$ 15 + 0.30 $\times$ 20

Expected sales (mean) = 15

Variance of the sales = 0.30 $\times$ [ 10 - 15 ] ² + 0.40 $\times$ [ 15 - 15 ] ² + 0.30 $\times$ [ 20 - 15 ] ²

Variance of the sales = 7.5 + 7.5

Variance of the sales = 15

Standard deviation =

V Variance

Standard deviation = $\sqrt{15}$ = 3.872983346

Coefficient of variation = 3.872983346 $\div$ 15

Coefficient of variation = 25.8%