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Are normally distributed with a mean of 68 inches and a standard deviation of 2 inches What is the probability that the height of a randomly selected female college basketball player is more than 66?
84% To solve this problem, you must first realize that 66 inches is one standard deviation below the mean. The empirical rule states that 34% will be between the mean and 1 standard deviation below the mean. We are looking for the prob. of the height being greater than 66 inches, which is then 50% (for the entire right side of the distribution) + 34%
How do you calculate probability given mean and standard deviation?
The mean and standard deviation do not, by themselves, provide enough information to calculate probability. You also need to know the distribution of the variable in question.
Is the uniform probability distribution's standard deviation proportional to the distribution's range?
no
What is the probability of scoring 115 when mean is 100 and standard deviation is 15?
The probability of scoring an exact value for a continuous variable is zero for any value. The probability of scoring 115 (or more) is 15.87%
If quartile deviation is 24. find mean deviation and standard deviation?
Information is not sufficient to find mean deviation and standard deviation.
What is the probability that the value of a randomly selected variable from a normal distribution will be more than 3 standard deviation from its mean value?
Approx 0.0027
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.
The life expectancy of Timely brand watches is normally distributed with a mean of four years and a standard deviation of eight months. a. What is the probability that a randomly selected watch will b?
To find the probability that a randomly selected Timely brand watch will last longer than a certain time, we first need to convert the time into the same unit as the standard deviation. Since the mean is 4 years (or 48 months) and the standard deviation is 8 months, we can use the z-score formula ( z = \frac{(X - \mu)}{\sigma} ) where ( X ) is the time in months, ( \mu ) is the mean (48 months), and ( \sigma ) is the standard deviation (8 months). After calculating the z-score, we can look up the corresponding probability in the standard normal distribution table to find the likelihood of a watch lasting longer than that specified time. Please provide the specific time in months or years for a complete answer.
A bank and loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. If an applicant is randomly selected find the probabili?
To find the probability of a randomly selected applicant receiving a credit rating above a certain value, you would first need to determine that value. For example, if you want to find the probability of an applicant having a rating above 250, you would calculate the z-score using the formula ( z = \frac{(X - \mu)}{\sigma} ), where ( X ) is the rating, ( \mu ) is the mean (200), and ( \sigma ) is the standard deviation (50). After calculating the z-score, you can use the standard normal distribution table or a calculator to find the corresponding probability.
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.
What is Normal of Standard Normal Probability Distribution?
with mean of and standard deviation of 1.
Are normally distributed with a mean of 68 inches and a standard deviation of 2 inches What is the probability that the height of a randomly selected female college basketball player is more than 66?
84% To solve this problem, you must first realize that 66 inches is one standard deviation below the mean. The empirical rule states that 34% will be between the mean and 1 standard deviation below the mean. We are looking for the prob. of the height being greater than 66 inches, which is then 50% (for the entire right side of the distribution) + 34%
The average amount customers at a certain grocery store spend yearly is 636.55 Assume the variable is normally distributed If the standard deviation is 89.46 find the probability that a randomly?
.820=82.0%
How do you calculate probability given mean and standard deviation?
The mean and standard deviation do not, by themselves, provide enough information to calculate probability. You also need to know the distribution of the variable in question.
The miles per gallon for a mid-size car is normally distributed with a mean of 32 and a standard deviation of 8 what is the probability that the mpg for a selected mid-size car would be less than 33.2?
The MPG (mileage per gallon) for a mid-size car is normally distributed with a mean of 32 and a standard deviation of .8. What is the probability that the MPG for a selected mid-size car would be less than 33.2?
Why is the standard normal probability distribution unique?
a mean of 1 and any standard deviation
What is Normal of probability distribution?
with mean and standard deviation . Once standardized, , the test statistic follows Standard Normal Probability Distribution.
