If p varies directly with T and p =105 when T=400.Find p when T =500
1 answer:
Answer:
<h3>p = 131.25</h3>Step-by-step explanation:
The variation p varies directly with T is written as
p = kT
where k is the constant of proportionality
To find p when T =500 we must first find the formula for the variation
That's
when p = 105 and T = 400
105 = 400k
Divide both sides by 400
<h3>So the formula for the variation is
<h2>when
T = 500
Substitute it into the above formula
That's
Simplify
The final answer is
<h3>p = 131.25</h3>Hope this helps you
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Answer:
Probability that the sample will have a mean that is greater than $52,000 is 0.0057.
Step-by-step explanation:
We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.
We select a random sample of 1,000 people.
<em>Let </em><em> = sample mean</em>
The z-score probability distribution for sample mean is given by;
Z = ~ N(0,1)
where, = population mean = $50,000
= population standard deviation = $25,000
n = sample of people = 1,000
The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.
So, probability that the sample will have a mean that is greater than $52,000 is given by = P( > $52,000)
P( > $52,000) = P(
>
) = P(Z > 2.53) = 1 - P(Z
2.53)
= 1 - 0.9943 = 0.0057
<em>Now, in the z table the P(Z </em><em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.</em>
Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.
The answers to the questions
Given Information:
Population mean = p = 60% = 0.60
Population size = N = 7400
Sample size = n = 50
Required Information:
Sample mean = μ = ?
standard deviation = σ = ?
Answer:
Sample mean = μ = 0.60
standard deviation = σ = 0.069
Step-by-step explanation:
We know from the central limit theorem, the sampling distribution is approximately normal as long as the expected number of successes and failures are equal or greater than 10
np ≥ 10
50*0.60 ≥ 10
30 ≥ 10 (satisfied)
n(1 - p) ≥ 10
50(1 - 0.60) ≥ 10
50(0.40) ≥ 10
20 ≥ 10 (satisfied)
The mean of the sampling distribution will be same as population mean that is
Sample mean = p = μ = 0.60
The standard deviation for this sampling distribution is given by
Where p is the population mean that is proportion of female students and n is the sample size.
Therefore, the standard deviation of the sampling distribution is 0.069.