Question

STATISTICS/PROBABILITY: Let X= the # of heads when 4 coins are tossed.

a.) Find the expected number of heads

b.) Find the variance and standard deviation

so far I have

x | 0 | 1 | 2 | 3 | 4 |

P(x) | 1/16 | 4/16 | 6/16 | 4/16 | 1/16 |

Answer #1

Solution:

The formula for expected number and variance and standard deviation are given as below:

Expected number = mean = ∑XP(X)

Variance = ∑ P(X)*(X - mean)^2

Standard deviation = Sqrt(Variance)

The calculation table is given as below:

X |
P(X) |
XP(X) |
P(X)*(X - mean)^2 |

0 |
0.0625 |
0 |
0.25 |

1 |
0.25 |
0.25 |
0.25 |

2 |
0.375 |
0.75 |
0 |

3 |
0.25 |
0.75 |
0.25 |

4 |
0.0625 |
0.25 |
0.25 |

Total |
1 |
2 |
1 |

Expected number = mean = ∑XP(X) = 2

**Expected number = 2**

Variance = ∑ P(X)*(X - mean)^2 = 1

**Variance = 1**

Standard deviation = Sqrt(Variance) = sqrt(1) = 1

**Standard deviation = 1**

Let X denote the number of heads than occur when four coins are
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independent and the probability of heads on each coin is 1/2,X is
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respectively. Do these results support the assumption that the
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A probability experiment consists of flipping 4 biased coins
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Complete the probability distribution for X
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X
P
(X = k)
P (X ≤
k)
0
1
0.4219
2
3
0.0469
4
0.0039
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E(X) =
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1.
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c) 20%
d)0
e) 1
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