Answer:
a) Expected sum of numbers obtained with 100 spins = 260
b) Standard deviation of the sum of numbers obtained with 100 spins = 12
c) Probability that the sum of numbers obtained on 100 spins is 245 to 255, inclusive, using the correction factor = 0.25344
Explanation:
Before solving for expected value and standard deviation of a 100 spins, we do for one spin first,
Expected value is given as
E(X) = Σ xᵢpᵢ
where xᵢ = each variable
pᵢ = probability of each variable
We have sample spaces 1, 2, and 4
with 1 appearing once, 2 appearing twice and 4 appearing twice too. So, their respective probabilities of turning up are (1/5), (2/5) and (2/5).
E(X) = (1)(1/5) + (2)(2/5) + (4)(2/5) = 2.6
Standard deviation = √(variance)
Variance = Var(X) = Σx²p − μ²
where μ = E(X) = 2.6
Σx²p = (1²)(1/5) + (2²)(2/5) + (4²)(2/5) = 8.2
Variance = 8.2 - 2.6² = 1.44
Standard deviation = √(variance) = √(1.44) = 1.2
Now for 100 spins,
E(100X) = 100 × E(X) = 100 × 2.6 = 260
Standard deviation (100X) = √100 × 1.2 = 10 × 1.2 = 12
Probability that the sum of numbers obtained on 100 spins is 245 to 255, inclusive, using the correction factor.
P(245 ≤ x ≤ 255)
Introducing the correction factor, we have
P(244.5 ≤ x ≤ 255.5) (correction factor is 0.5)
Using normal distribution tables, we first standardize 244.5 and 255.5.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
Note that the mean = μ = 260
Standard deviation = σ = 12
For 244.5
z = (x - μ)/σ = (244.5 - 260)/12 = - 1.29
For 255.5
z = (x - μ)/σ = (255.5 - 260)/12 = - 0.38
To determine the required probability
P(244.5 ≤ x ≤ 255.5) = P(-1.29 ≤ z ≤ -0.38)
We'll use data from the normal probability table for these probabilities
P(244.5 ≤ x ≤ 255.5) = P(-1.29 ≤ z ≤ -0.38)
= P(z ≤ -0.38) - P(z ≤ -1.29)
= 0.35197 - 0.09853
= 0.25344
Hope this Helps!!!