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The Normal Probability Curve


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Strings (SiPjAjk) = S7P6A64     Base Sequence = 12735     String Sequence = 12735 - 6 - 64

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The Normal Probability Curve

(a) Suppose that the frequencies of some data is normally distributed and figure 118.5 represents the probability curve. What is the probability of a value occurring between a and b?
(b) The weight of a large number of grapefruits were found to be normally distributed with a mean of 1 lb and a standard deviation of 3 oz. What is the probability that any one grapefruit has a weight between 1 lb 3 oz and 1 lb 6 oz?
(c) The average number of persons joining a certain queue in one minute is 2. What is the probability that 5 persons will join the queue in one minute?

The strings: S7P6A64 (Grouping - Multi-criteria).

The math:
Pj Problem of Interest is of type grouping (multi-criteria). Grouping is at the heart of statistics. The grouping may be permutational or combinational, single criterion or multi-criteria grouping.

The Normal Probability Curve

(a) The probability that a value lies within a and b is given by the area under the curve between a and b
So, The probability that a value lies within a and b = ∫a b f(x) dx.
Where f(x) is as indicated in figure 118.5
Mean = arithmetic average = [f1x1 + f2x2 ... fnxn]/n
Where fi is the number of xi in the data. i = 1, 2,...n.
Standard deviation σ = [Σi [fi(xi - mean)2]/n]1/2. i = 1, 2,...n.

(b) Normal frequency curves are varied. However, they are all characterized by their mean and standard deviation. Irrespective of the value of the mean and standard deviation, 68.2 % of the data lie within one standard deviation (σ) on either side of the mean; 95.4 % of the data lie within 2σ on either side of the mean and 99.8 % lie within 3σ on either side of the mean.
The interval of interest is between σ and 2σ to the right of the mean.
So, desired probability = 0.954/2 - 0.682/2 = 0.477 - 0.341 = 0.136.

(c) The Poisson distribution is the pertinent probability distribution of interest
The Poisson distribution says that if the mean number of events of a particular type in a fixed time interval is μ, then the probability of n events p(n) occurring in one interval is given by:
p(n) = (μne)/ n! Where n! is n factorial = n(n-1)(n-2)...1
So, p(5) = (25e-2)/(5x4x3x2x1) = [32/(2.718)2]/120 = 0.036.

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