Question

x is a normally distributed random variable with mean 46 and standard deviation 16.
what is the probability that x is between 16 and 76?
write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Standardize the values

The formula for standardization is \( z=\frac{x - \mu}{\sigma} \), where \( \mu = 46 \), \( \sigma=16 \).
For \( x = 16 \): \( z_1=\frac{16 - 46}{16}=\frac{- 30}{16}=-1.875 \)
For \( x = 76 \): \( z_2=\frac{76 - 46}{16}=\frac{30}{16}=1.875 \)

Step2: Find the probabilities

We want to find \( P(16<X<76)=P(-1.875<Z<1.875) \)
Using the property \( P(-a < Z < a)=2\Phi(a)-1 \), where \( \Phi(a) \) is the cumulative distribution function of the standard - normal distribution.
Looking up \( \Phi(1.875) \) in the standard - normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: normalcdf\((-1.875,1.875,0,1)\)), we know that \( \Phi(1.875)\approx0.9696 \)
Then \( P(-1.875 < Z < 1.875)=2\times0.9696-1 \)
\(=1.9392 - 1=0.9392\)

Answer:

\(0.939\)