Q: The average gas mileage of a certain model car is 28 miles per gallon If the gas mileages are normally distributed with a standard deviation of 1.7 find the probability that a car has a gas mileage?

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A particular fruit's weights are normally distributed, with a mean of 760 grams and a standard deviation of 15 grams. If you pick one fruit at random, what is the probability that it will weigh between 722 grams and 746 grams-----A particular fruit's weights are normally distributed, with a mean of 567 grams and a standard deviation of 25 grams.

.820=82.0%

z score = (400-500)/80=-1.25 probability under curve to right of -1.25 = .8944

68.2%

The answer is 0.1586

Related questions

A particular fruit's weights are normally distributed, with a mean of 760 grams and a standard deviation of 15 grams. If you pick one fruit at random, what is the probability that it will weigh between 722 grams and 746 grams-----A particular fruit's weights are normally distributed, with a mean of 567 grams and a standard deviation of 25 grams.

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?

.820=82.0%

The mean and standard deviation. If the data really are normally distributed, all other statistics are redundant.

Since adult cereal is normally distributed with known mean and known standard deviation, the best way to find this probability is to find the z-score corresponding to adult cereal weighing 895 grams and then using a normal probability table to find an approximate probability (usually accurate to 4 decimal places). The z-score is (895-920)/10=-2.5. So, the probability that adult cereal will weigh less than 895 grams is 0.0062.

There are many variables that are not normally distributed. You can describe them using a probability distribution function or its cumulative version; you can present them graphically.

Anything that is normally distributed has certain properties. One is that the bulk of scores will be near the mean and the farther from the mean you are, the less common the score. Specifically, about 68% of anything that is normally distributed falls within one standard deviation of the mean. That means that 68% of IQ scores fall between 85 and 115 (the mean being 100 and standard deviation being 15) AND 68% of adult male heights fall between 65 and 75 inches (the mean being 70 and I am estimating a standard deviation of 5). Basically, even though the means and standard deviations change, something that is normally distributed will keep these probabilities (relative to the mean and standard deviation). By standardizing these numbers (changing the mean to 0 and the standard deviation to 1) we can use one table to find the probabilities for anything that is normally distributed.

z score = (400-500)/80=-1.25 probability under curve to right of -1.25 = .8944

68.2%

The answer is 0.1586

If you have a variable X distributed with mean m and standard deviation s, then the z-score is (x - m)/s. If X is normally distributed, or is the mean of a random sample then Z has a Standard Normal distribution: that is, a Gaussian distribution with mean 0 and variance 1. The probability density function of Z is tabulated so that you can check the probability of observing a value as much or more extreme.

mean = 11ftstandard deviation = 1.6ftx=13z= 13-11 = 1.251.6P(0