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If the amount of soft drink that I consume on any given day is independent of consumption on any other day and is normally distributed with\(\mu = 13\)oz and\(\sigma = 2\)and if I currently have two six-packs of\({\bf{16}}\)-oz bottles, what is the probability that I still have some soft drink left at the end of\({\bf{2}}\)weeks?

Short Answer

Expert verified

The probability that I still have some soft drink left at the end of two weeks is \(0.9099\).

Step by step solution

01

Definition of Probability

The state of being able to do something is known as probability. It's a branch of mathematics concerned with event randomness. The value is indicated from zero to one. The idea of probability was developed in mathematics to predict the possibility of events occurring. Probability is simply the probability of anything happening

02

Calculation for the determination of probability.

The random variable of interest (total consumption of soft drinks) is

\({T_0} = {X_1} + {X_2} + \ldots + {X_{14}}\)

where each \({X_i},i = 1,2, \ldots ,14\)are random variable with normal distribution with given parameters (consumption of soft drinks on day i).

In order to compute the requested probability, mean value and standard deviation are required (to standardize \({T_0}\)and obtain standard normal distribution).

The mean value of the random variable \({T_0}\)is

\(\begin{aligned} \mu _{{T_0}}} &= E\left( {{X_1} + {X_2} + \ldots + {X_{14}}} \right &= E\left( {{X_1}} \right) + E\left( {{X_2}} \right) + \ldots + E\left( {{X_{14}}} \right)\\ &= 14 \cdot \mu \\ &= 14 \cdot 13\\ &= 182\end{aligned}\)

The variance of random variable \({T_0}\)is

\(\begin{aligned}\sigma _{{T_0}}^2 = V\left( {{T_0}} \right) &= V\left( {{X_1} + {X_2} + \ldots + {X_{14}}} \right) &= V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + \ldots + V\left( {{X_{14}}} \right)\\ &= 14 \cdot {\sigma ^2}\\ &= 14 \cdot 4\\ &= 56\end{aligned}\)

03

Calculation for the determination of probability.

\(\begin{array}{c}\sigma _{{T_0}}^2 = V\left( {{T_0}} \right) = V\left( {{X_1} + {X_2} + \ldots + {X_{14}}} \right) = V\left( {{X_1}} \right) + V\left( {{X_2}} \right) + \ldots + V\left( {{X_{14}}} \right)\\ = 14 \cdot {\sigma ^2}\\ = 14 \cdot 4\end{array}\)

The standard deviation of random variable \({T_0}\)is

\({\sigma _{{T_0}}} = \sqrt {56} = 7.483\)

The probability that some drinks are left at the end of \(14\) days, where total number of drinks is \(2 \cdot 6 \cdot 16 = 192\)(two six-packs, \(16\)bottles), is

\(\begin{array}{c}P\left( {{T_0} < 192} \right) = P\left( {\frac{{{T_0} - {\mu _{{T_0}}}}}{{{\sigma _{{T_0}}}}} < \frac{{192 - 182}}{{7.483}}} \right)\\ = P(Z < 1.34)\mathop = \limits^{(1)} 0.9099,\end{array}\)

(1): from the normal probability table in the appendix. The probability can also be computed with software.

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