In order to determine this probability, we calculate using this difference:
To obtain these probabilities, it’s possible to utilize normal standard distribution tables, a calculator, or software like Excel. The accompanying figure displays the results achieved. Here’s a detailed breakdown of the steps: Relevant concepts include the normal distribution, which describes a probability distribution that is symmetric regarding the mean, demonstrating that occurrences close to the mean are more likely than those farther away. The Z-score represents a statistical measure illustrating how far a value is from the average of a set, expressed in standard deviations.
For our analysis, let X denote the random variable representing weights in a population, with its distribution characterized by:
We’re specifically interested in this probability. The most effective approach to address this issue is through the standard normal distribution and the Z-score calculation, expressed as:
Applying this formula to our probability provides the following:
This allows us to calculate this probability with the provided difference:
We use standard distribution tables, a calculator, or Excel for determining these probabilities. The graph illustrates the resulting outcome.
This problem can be addressed by applying the normal approximation to a binomial distribution.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
(5r - 4)(r² - 6r + 4)
uses the distributive property for multiplication.
This expands to 5r(r² - 6r + 4) - 4(r² - 6r + 4)
which results in 5r³ - 30r² + 20r - 4r² + 24r - 16
as you combine like terms and simplify.
The outcome is 5r³ - 30r² - 4r² + 20r + 24r - 16
leading to a final expression of 5r³ - 34r² + 44r - 16. Choice A.
A. Mean and standard deviation.
The sampling distribution’s mean closely matches the population mean. Since the population mean is 174.5, the sampling distribution’s mean equals this value.
The standard deviation of the sampling distribution is:
σₓ̄ = σ / √n
Plugging in values:
σₓ̄ = 6.9 / √25 = 1.38
b. Calculate z-scores for both values:
z = (value - mean) / standard deviation
For 172.5:
z = (172.5 - 174.5) / 1.38 = -1.49
Corresponding probability ≈ 0.068
For 175.8:
z = (175.8 - 174.5) / 1.38 = 0.94
Corresponding probability ≈ 0.83
The difference between these probabilities is 0.762.
Approximately 0.762 × 200 = 152 sample means lie between 172.5 and 175.8.
c. Z-score for 172 cm:
z = (172 - 174.5) / 1.38 = -1.81
Probability ≈ 0.03
Hence, samples with means below 172 cm equal 0.03 × 200 = 6.
X² - 7x - 8 = 0
Applying the Viet theorem
x₁ + x₂ = -p
-1 + x₂ = 7
x₂ = 7 + 1
x₂ = 8
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