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A car hiring firm has two cars. The demand for cars on each day is distributed as a Poisson variate, with mean 1.5. Calculate the proportion of days on which neither car is used - Business Mathematics and Statistics

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प्रश्न

A car hiring firm has two cars. The demand for cars on each day is distributed as a Poison variate, with mean 1.5. Calculate the proportion of days on which neither car is used

बेरीज

उत्तर

In a posion distribution n=2

Mean λ = 1.5

x follows poison distribution

With in P(x) = `("e"^(-lambda) lambda^x)/(x!)`

P(neither car is used) = P(X = 0)

`("e"^(-1.5) (1.5)^0)/(0!) = "e"^(-1.5)`

= 0.2231

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Probability Distributions - Exercise 7.2 [पृष्ठ १६०]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
पाठ 7 Probability Distributions
Exercise 7.2 | Q 8. (i) | पृष्ठ १६०

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