use the appropriate normal distribution to approximate the resulting binomial distributions. a fair coin is tossed 130 times. what is the probability of obtaining between 55 and 69 tails, inclusive? a) 0.7523 b) 0.7134 c) 0.7221 d) 0.7179 e) 0.7494 f) none of the above.

Standard Deviation is a measure that shows what quantity variation (such as unfold, dispersion, spread,) from the mean exists. the quality deviation indicates a “typical” deviation from the mean. it's a well-liked live of variability as a result of it returns to the first units of live of the info set.

Main body:

X - Binomial(n = 130, p = 0.5)

Binomial can be approximated to normal with: μ = np = 130*0.5 = 65

σ = √np(1 − p) - = √130* (0.5)(1 − 0.5)

σ = 5.7

P(55 ≤ X ≤ 69 )

Since we are approximating a discrete binomial distribution by continuous normal distribution, values between 54.5 and 68.5 both approximate to 55. Thus, "between 55 and 69" corresponds to continuous normal distribution with P(54.5 < X 68.5) after continuity correction.

Normal Distribution, x₁ = 54.5, x2 = 68.5, µ = 65, σ = 5.7

We convert this to standard normal using z =

z₁ = 54.5-65/ 5.7 ≈ - 1.84

z₂ = 68.5-65/ 5.7 ≈ 0.614

P(54.5 < X < 68.5) = P(- 1.84 < Z < 0.614)

= P(Z < 0.614) − P(Z <- 1.84)

= 0.7291 - 0.0329 (from z-table)

= 0.6962

Hence the probability is 0.6962

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