A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, what is the probability that a head will show between 30 and 36 times inclusive?

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A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, what is the probability that a head will show between 30 and 36 times inclusive?

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Answer 1 · 2017-12-02T14:27:21

$0.505$ $0.505$

Explanation:

$n = 60 p = \frac{1}{2} \Rightarrow m = n p = 30$ $n = 60 " " p = 1/2 => m = n p = 30 " "$
$and$ $"and"$
$s = \sqrt{n p (1 - p)} = \sqrt{15}$ $s = sqrt(np(1-p)) = sqrt(15) " "$

$z 2 = \frac{36.5 - 30}{\sqrt{15}} = 1.678$ $z2 = (36.5 - 30)/sqrt(15) = 1.678$
$z 1 = \frac{29.5 - 30}{\sqrt{15}} = - 0.129$ $z1 = (29.5 - 30)/sqrt(15) = -0.129$

$We search z1 and z2 in a table for z-values and get$ $"We search z1 and z2 in a table for z-values and get "$
$0.953 - 0.448 = 0.505$ $0.953 - 0.448 = 0.505$

$Remark : we use 36.5 and 29.5 instead of 36 and 30$ $"Remark : we use 36.5 and 29.5 instead of 36 and 30"$
$for reason of a continuity correction.$ $"for reason of a continuity correction."$

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A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, what is the probability that a head will show between 30 and 36 times inclusive?

1 Answer

Explanation:

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