Question

Costs for standard veterinary services at a local animal hospital follow a normal distribution with a mean of $83 and a standard deviation of $21. what is the probability that one bill for veterinary services costs between $51 and $114?

Answer 1

The probability that a randomly selected data from a normally distributed dataset with mean of μ, and standard deviation of σ, is between two values a and b is given by:

$P(a\ \textless \ X\ \textless \ b)=P(X\ \textless \ b)-P(X\ \textless \ a) \\ \\ =P\left(z\ \textless \ \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$

Given that the costs for standard veterinary services at a local animal hospital follow a normal distribution with a mean of $83 and a standard deviation of $21.

The probability that one bill for veterinary services costs between $51 and $114 is given by:

$P(51\ \textless \ X\ \textless \ 114)=P\left(z\ \textless \ \frac{114-83}{21} \right)-P\left(z\ \textless \ \frac{51-83}{21} \right) \\ \\ =P(z\ \textless \ 1.476)-P(z\ \textless \ -1.524)=0.93005-0.06378 \\ \\ =\bold{0.8663}$