हिंदी

For the following probability distribution: X – 4 – 3 – 2 – 1 0 P(X) 0.1 0.2 0.3 0.2 0.2 E(X) is equal to ______. - Mathematics

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प्रश्न

For the following probability distribution:

X – 4 – 3 – 2 – 1 0
P(X) 0.1 0.2 0.3 0.2 0.2

E(X) is equal to ______.

विकल्प

  • 0

  • – 1

  • – 2

  • – 1.8

MCQ
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उत्तर

For the following probability distribution:

X – 4 – 3 – 2 – 1 0
P(X) 0.1 0.2 0.3 0.2 0.2

E(X) is equal to – 1.8.

Explanation:

We know that

E(X) = `sum_("i" = 1)^"n" "X"_"i""P"_"i"`

= (– 4)(0.1) + (– 3)(0.2) + (– 2)(0.3) + (– 1)(0.2) + 0(0.2)

= – 0.4 – 0.6 – 0.6 – 0.2

= – 1.8

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अध्याय 13: Probability - Exercise [पृष्ठ २८४]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 13 Probability
Exercise | Q 88 | पृष्ठ २८४
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