The weight of a medium-sized orange selected at random from a large bin of oranges at a local supermarket is a normally distributed random variable with mean μ = 12 ounces and standard deviation σ = 1.2 ounces. suppose we independently select two oranges at random from the bin. what is the probability that the difference in the weights of the two oranges exceeds 3 ounces?

0.0392

Step-by-step explanation:

The weight is normally distributed with :

Mean (\mu)= 12 ounces

Standard deviation (\sigma ) = 1.2 ounces

Let X_1 be the weight of first orange and X_2 be the weight of second orange.

Let X be the difference of the weight of two oranges (X_1-X_2)

X_1(12,1.2^{2}) and X_2(12,1.2^{2})

X(12-12,√1.2^{2}+1.2^{2}) = X(0,1.6971)

We need to calculate P(X>3) = 1-P(X≤3)

For calculating P(X≤3) in normal distribution, we use

Z =

 =  

 = 1.7677

P(Z≤ 1.7677) = 0.9608 (From normal distribution table)

P(X > 3) = 1-0.9608

            = 0.0392

$450.00 for 6 weeks

or

$300

step-by-step explanation:

$200.00 for first week $50.00 x5 equals $250 plus $200 equals $450.00 for six weeks.

$200 for first week

$200 x 0.10 = $20

$20 x 5 = $100

$200 + $100 = $300

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step-by-step explanation: