Question

the nutbuddy company sells a 1 kg format bag of mixed nuts that contains peanuts, almonds, brazil nuts and cashews. nutbuddy claims that the mixture contains an equal amount by weight of each nut, but in reality the amount of each nut is a normally distributed random variable, with mean 250 g and with standard deviation 15 g for peanuts, 18.5 g for almonds, 22.5 g for brazil nuts, and 23.5 g for cashews. the amounts of each type of nut are mutually independent. if nutbuddy buys its peanuts for 5.00 $/kg, its almonds for 15.00 $/ kg, its Brazil nuts for 7.15 $/kg, and its cashews for 20.15 $/kg, what is the probability that a randomly selected 1 kg bag of the nut mixture contains between $11.31 and $12.51 worth of nuts?

Answer 1

The distributions are following for each type of nut:

Peanuts~ N(250, 15)

Almonds~ N(250, 18.5)

Brazil nuts~ N(250, 22.5)

Cashews~ N(250, 23.5)

Nuts worth = 0.005*N(250, 15) + 0.015*N(250, 18.5) + 0.00715*N(250, 22.5) + 0.02015*N(250, 23.5)

It will be a normal distribution with the following statistics:

Mean = 250*(0.005 + 0.015 + 0.00715 + 0.02015) = 250*(0.0473) = 11.825

Variance = Sum of individual variances = (0.005*15)² + (0.015*18.5)² + (0.00715*22.5)² + (0.02015*23.5)² = 0.333

Hence, Nuts worth~ N(11.825, 0.333)

We need to find: P(11.31 < X < 12.51) = P( (11.31 - 11.825)/0.333 < z < (12.51 - 11.825)/0.333)

= P(-1.55 < z < 2.06) = P(z < 2.06) - P(z < -1.55)

From the z-table,

= 0.9803 - 0.0606 = 0.9197

The probability that a randomly selected 1 kg bag of the nut mixture contains between $11.31 and $12.51 worth of nuts is 0.9197