The Normal Probability Curve
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The Normal Probability Curve

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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