First, determine how many standard deviations above the mean one would have to be to be in the 75th percentile. This can be found by using a z table and finding the z associated with 0.75. The value of z is 0.674. Thus, one must be .674 standard deviations above the mean to be in the 75th percentile

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3 years ago

Costs for standard veterinary services at a local animal hospital follow a Normal distribution with a mean of $72 and a standard

Romashka [77]

Answer:

$P(32$

And we can find this probability with this difference:

$P(-1.905$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-1.905$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the costs of services of a population, and for this case we know the distribution for X is given by:

$X \sim N(72,21)$