# Statistics Assignment - Speech or Presentation Example

Summary
The condition occurs in both conditions. Taking this into consideration, the total probability of a customer waiting at least 12 minutes or between 8 and…

## Extract of sample "Statistics Assignment"

Download file to see previous pages The probability of winning the lottery is then around nine in a billion. It should be noted that the resulting equation actually simplifies to a multiplication rule. However, the binomial distribution may be used to model more complex lottery options.
In the construction of a 5-digit number, the order in which is numbers are drawn is clearly important. Also, since repetition of digits is not allowed, a single digit chosen is removed from the pool of possible succeeding digits. These conditions indicate that the total number of digits can simply be taken using a permutation.
The given table can only be classified as a probability distribution if its individual probabilities are between zero and one and the total probability is equal to one. In the given table, the probability at X = 2 is greater than one. Since no negative probabilities are present, the total probability is also greater than one. As both requirements are not met, the table cannot be a probability distribution.
The sample data is relatively sparse (n < 30) which warrants the use of the t-distribution instead of z-scores. To verify whether or not the population mean lifetimes approximate that of the advertized mean, it is necessary to obtain the sample mean and standard deviation. To aid in this, the following table is constructed:
To determine whether a relationship exists between the two variables, the given data may be fitted onto a function. The simplest approach is to use a linear model for approximating the relation. To aid in this, a column representation may be constructed. It should be noted that the independent variable based on the description is the cost of the product.
An assumed frequency distribution is assumed in this case which allows the use of the Goodness of Fit test. The computations for the chi square statistic can then be determined as shown in the following table:
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