Answer:
a) P(48 < x < 58) = 0.576
b) P(X ≥ 1) = 0.9863
c) E(X) 2.88
d) P(x < 48) = 0.212
e) P(X > 2) = 0.06755
Step-by-step explanation:
The mean of the wingspan of the birds = μ = 53.0 mm
The standard deviation = σ = 6.25 mm
a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.
To solve this, we first normalize/standardize the two wingspans concerned.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ
For wingspan 48 mm
z = (48 - 53)/6.25 = - 0.80
For wingspan 58 mm
z = (58 - 53)/6.25 = 0.80
To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)
We'll use data from the normal probability table for these probabilities
P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576
b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)
P(X ≥ 1) = 1 - P(X=0)
Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = number of birds = 5
x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0
p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576
q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424
P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137
The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863
c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.
Expected value is a sum of each variable and its probability,
E(X) = mean = np = 5×0.576 = 2.88
d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long
Using the normal distribution tables again
P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212
e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)
This is also a binomial distribution problem,
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = number of birds = 5
x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.
p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212
q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788
P(X > 2) = P(X=3) + P(X=4) + P(X=5)
P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218
P(X > 2) = 0.06755122607 = 0.06755
