Respuesta :
Given a normal random distribution with mean μ and standard deviation σ:
[tex]\begin{gathered} \mu=121 \\ \sigma=23 \end{gathered}[/tex]The z-score can be calculated using the formula:
[tex]Z=\frac{X-\mu}{\sigma}[/tex]Where X is a value of the normal random distribution. We need the probability that a randomly selected utility bill is between $115 and $130. The z-scores of these values are:
[tex]\begin{gathered} Z_1=\frac{115-121}{23}=-\frac{6}{23}\approx-0.26087 \\ Z_2=\frac{130-121}{23}=\frac{9}{23}\approx0.39130 \end{gathered}[/tex]Then, we need to calculate the probability of:
[tex]P(Z_1\leq Z\leq Z_2)=P(-0.26087\leq Z\leq0.39130)[/tex]Using the tables of the z-distribution, this probability is:
[tex]P(-0.26087\leq Z\leq0.39130)\approx0.255[/tex]