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For a binomial probability distribution, the variance is n*p*q which is 80*.3*.7 = 16.8. The standard deviation is square root of the variance which is 4.099; rounded is 4.1. The mean for a binomial probability distribution is n*p or 80*.3 or 24.

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What is the p-value for the following hypothesis test. H0 M equals 23 H1 M not 23 n equals 50 x-bar equals 21.25 standard deviation equals 5?

okay wikianswers messed up the question. H0: M=23; H1: M not 23; n=50; x-bar = 21.25; standard deviation = 5. please help!


What is the mean and standard deviation of 60 percent of 300?

There is insufficient information in the question to answer it. In order to compute a mean and a standard deviation, you need at least two data points, but the question only gave one. Please restate the question.


What is the z-score for 45.00?

There is insufficient information in the question to properly answer it. In order to compute a Z score from a raw score, you need the mean and the the standard deviation, neither of which was given. Please restate the question.


What that women's heights are normally distributed with a mean given by mu equals 64.1 and a standard deviation given by sigma equals 3.2 in . If 1 woman is randomly selected find the probability th?

To find the probability of a randomly selected woman's height, we can use the properties of the normal distribution. If women's heights are normally distributed with a mean (μ) of 64.1 inches and a standard deviation (σ) of 3.2 inches, we can calculate probabilities for specific ranges of heights using the Z-score formula: ( Z = \frac{(X - μ)}{σ} ). For a specific height or range, we can then look up the corresponding probability in standard normal distribution tables or use statistical software. Please specify the height or range you are interested in for a more precise calculation.


Assume that women's heights are normally distributed with a mean given by mu equals 63.6 in and a standard deviation given by sigma equals 2.1 in (a) If 1 woman is randomly selected find the probabili?

To find the probability of a randomly selected woman having a height within a specific range, we can use the normal distribution with the given mean (μ = 63.6 inches) and standard deviation (σ = 2.1 inches). For instance, if we want to find the probability that a randomly selected woman is shorter than 65 inches, we would calculate the z-score using the formula ( z = \frac{(X - \mu)}{\sigma} ), where ( X ) is the height in question. After calculating the z-score, we would consult the standard normal distribution table or use a calculator to find the corresponding probability. If you have a specific height range in mind, please specify for a more detailed calculation.

Related Questions

Can standard deviation be a negative?

No. Standard deviation is the square root of a non-negative number (the variance) and as such has to be at least zero. Please see the related links for a definition of standard deviation and some examples.


Is it true that the standard deviation is equal to the square root of the variance?

Yes. Please see the related link, below.


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The cumulative probability up to the mean plus 1 standard deviation for a Normal distribution - not any distribution - is 84%. The reference is any table (or on-line version) of z-scores for the standard normal distribution.


What is the p-value for the following hypothesis test. H0 M equals 23 H1 M not 23 n equals 50 x-bar equals 21.25 standard deviation equals 5?

okay wikianswers messed up the question. H0: M=23; H1: M not 23; n=50; x-bar = 21.25; standard deviation = 5. please help!


What is the mean and standard deviation of 60 percent of 300?

There is insufficient information in the question to answer it. In order to compute a mean and a standard deviation, you need at least two data points, but the question only gave one. Please restate the question.


Who hdhdhdhfhfhfhhfhdjjdjsj iausjsjjshrrhfhththfjkjjhdose jsjsyeyehehrhrhfhfnnshcshereyjykirrtssryjj am ttadfiuytgghj dispersion?

It seems like your question contains a lot of random characters and may not be clear. If you're asking about "dispersion," it generally refers to the way in which data points are spread out or distributed in a dataset. This can involve concepts like variance, standard deviation, and range, which help to understand the variability within a set of values. Please clarify if you meant something specific!


What is the z score for 0.71?

This question cannot be answered. You need the mean and standard deviation in order to compute a Z score for a Raw score. Please restate the question.


What z-score corresponds to this value?

There is insufficient information in the question to answer it. To determine Z score, you need raw score, mean, and standard deviation. Please restate the question.


What is the z-score for a test score of 110?

In order to know the z-score, given a test score, you must also know the mean and the standard deviation. Please restate the question.


What is the z-score for 45.00?

There is insufficient information in the question to properly answer it. In order to compute a Z score from a raw score, you need the mean and the the standard deviation, neither of which was given. Please restate the question.


How many teaspoon equals 1 c?

A "c" is not a standard measure that we know. -please re-write when you know the measure.


What that women's heights are normally distributed with a mean given by mu equals 64.1 and a standard deviation given by sigma equals 3.2 in . If 1 woman is randomly selected find the probability th?

To find the probability of a randomly selected woman's height, we can use the properties of the normal distribution. If women's heights are normally distributed with a mean (μ) of 64.1 inches and a standard deviation (σ) of 3.2 inches, we can calculate probabilities for specific ranges of heights using the Z-score formula: ( Z = \frac{(X - μ)}{σ} ). For a specific height or range, we can then look up the corresponding probability in standard normal distribution tables or use statistical software. Please specify the height or range you are interested in for a more precise calculation.