A random sample of n = 83 measurements is drawn from a binomial population with probability of success 0.4. Complete parts a thr
1 answer:
Answer:
a) The mean of the sampling distribution of p is 0.4.
The standard deviation of the sampling distribution of p is 0.0538.
b) The shape of the sampling distribution of p is approximately normal because the sample size is large.
c) z=0.19
Step-by-step explanation:
We have a random sample of size n=83, drawn from a binomial population with proabiliity p=0.4. We have to compute the characteristic of the sampling distribution of the sample proportion.
a) The mean of the sampling distribution is equal to the mean of the distribution p:
The standard deviation of the sampling distribution is:
b) The shape of sampling distribution with enough sample size tend to be approximately normal. In this case, n=83 is big enough for a binomial distribution.
c) The z-score fof p-0.41 can be calculated as:
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Answer:
The heaviest 13% of fruits will weight more than 590grs.
Step-by-step explanation:
Hello!
You know that X: the weight of a fruit, in grams, has a normal distribution with mean μ= 549gr and standard deviation σ= 36gr.
If you were to graph this distribution it will be bell-shaped and the area under that curve represents 100% of the distribution of the population of the fruit's weights. Where the most common weights will be found near the center of the bell (μ), the lightest fruits will be in the left tail of the curve and the heaviest fruits will be in the right tail of the curve.
Then the value that indicates the heaviest 13% of the weight of the fruit will be on the right tail of the curve (and will be greater than the population mean)
If above this value, let's call it x₀, you find 13% of the distribution, then is logical to think that below it you'll find the rest of the distribution 100% - 13% )= 87%, symbolically:
P(X≥x₀)=0.13 ⇒ P(X<x₀)= 0.87
See the first attachment.
To find this value is best to work using the standard normal distribution since the standard normal distribution is tabulated. Any value of any random variable X with normal distribution can be "converted" by subtracting the variable from its mean and dividing it by its standard deviation.
So for our mystery value x₀, there is a counterpart value under the standard normal distribution, let's call it z₀, which also separates the bottom 87% of the distribution from the top 13%.
P(Z≥z₀)=0.13 ⇒ P(Z<z₀)= 0.87
The table shows you values of probability P(Z<z)= 1 - α So is best if you use the expression P(Z<z₀)= 0.87 to find the corresponding Z value. Using the right entry of the Z-table, you have to locate the value of probability in the body od the table and then reach the margins to find the value of Z. Remember, the integer and first digit of the value are on the first column and the second digit of the value is found in the first row of the table. See the second attachment.
z₀= 1.13
Now that you know the value under the standard normal distribution you have to transform it to a value of the study variable using:
Z= (X - μ)/σ~N(0;1)
z₀= (x₀ - μ)/σ
z₀*σ= x₀ - μ
x₀= (z₀ * σ) + μ
x₀= (1.13 * 36) + 549
x₀= 589.68 ≅ 590grs
I hope this helps!
