Mount Kilimanjaro is the highest mountain peak in Africa with an elevation of 19,341 feet. What is the
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This is a typical binomial distribution since we are dealing with only two possible outcomes. For this distribution we have:
n = population = 1400
p = probability of positive answer = 0.44
A) The mean of a binomial distribution is
μ = np
= 1400 × 0.44
= 616
Hence, <span>μ = 616
B) The standard deviation of a binomial distribution is
</span>σ = √[np(1 - p)]
= √[1400×0.44×(1-0.44)]
= √(1400×0.44×0.56)
= 18.6
Hence, σ = 18.6
C) A binomial distribution B(n, p) can be approximated by a normal distribution N(μ, σ) only when:
np ≥ 10 (we have 616 > 10)
np(1 - p) ≥ 10 (we have 344.96 > 10)
Hence, the normal approximation is N(616, 18.6)
D) We need to calculate the probability of the sample population to be within 3% of the true mean.
In order to do so, as a first thing, we need to calculate the sample standard deviation, which will be given by the formula:
= √[0.44 × (1 - 0.44) ÷ 1400]
= √0.000176
= 0.013
Now, we need to calculate the z-score associated with the values we want: since we want 3% from the mean, it means between 41% and 47%.
The z-score can be calculated by the formula:
z = (Y - p) / σ
z₁ = (0.41 - 0.44) / 0.013 = -2.31
z₂ = (0.47 - 0.44) / 0.013 = 2.31
Therefore, we can write
P(0.41 ≤ Y ≤ 0.47) = P(-2.31 ≤ z ≤ 2.31)
= P(z ≤ 2.31) - P(z ≤ -2.31)
= P(z <span>≤ 2.31) - [1 - P(z <span>≤ 2.31)]
= 2×P(</span>z </span>≤ 2.31) - 1
From a standard normal distribution table, we find:
P(z <span>≤ 2.31) = 0.9896
Therefore:</span>
<span>2×P(z </span><span>≤ 2.31) - 1 = 2 </span>× 0.9896 - 1
= 0.9792
Hence, the probability to find a proportion within 3% of the real mean is 97.92%
n = population = 1400
p = probability of positive answer = 0.44
A) The mean of a binomial distribution is
μ = np
= 1400 × 0.44
= 616
Hence, <span>μ = 616
B) The standard deviation of a binomial distribution is
</span>σ = √[np(1 - p)]
= √[1400×0.44×(1-0.44)]
= √(1400×0.44×0.56)
= 18.6
Hence, σ = 18.6
C) A binomial distribution B(n, p) can be approximated by a normal distribution N(μ, σ) only when:
np ≥ 10 (we have 616 > 10)
np(1 - p) ≥ 10 (we have 344.96 > 10)
Hence, the normal approximation is N(616, 18.6)
D) We need to calculate the probability of the sample population to be within 3% of the true mean.
In order to do so, as a first thing, we need to calculate the sample standard deviation, which will be given by the formula:
= √[0.44 × (1 - 0.44) ÷ 1400]
= √0.000176
= 0.013
Now, we need to calculate the z-score associated with the values we want: since we want 3% from the mean, it means between 41% and 47%.
The z-score can be calculated by the formula:
z = (Y - p) / σ
z₁ = (0.41 - 0.44) / 0.013 = -2.31
z₂ = (0.47 - 0.44) / 0.013 = 2.31
Therefore, we can write
P(0.41 ≤ Y ≤ 0.47) = P(-2.31 ≤ z ≤ 2.31)
= P(z ≤ 2.31) - P(z ≤ -2.31)
= P(z <span>≤ 2.31) - [1 - P(z <span>≤ 2.31)]
= 2×P(</span>z </span>≤ 2.31) - 1
From a standard normal distribution table, we find:
P(z <span>≤ 2.31) = 0.9896
Therefore:</span>
<span>2×P(z </span><span>≤ 2.31) - 1 = 2 </span>× 0.9896 - 1
= 0.9792
Hence, the probability to find a proportion within 3% of the real mean is 97.92%