Answer:
required probability is 0.8413
Step-by-step explanation: Given mean [tex]\mu =15[/tex] minutes, standard deviation [tex]\sigma =4[/tex] seconds, we need to find [tex]P\left ( X\leq 19 \right )[/tex]. Using z score
[tex]z=\frac{X-\mu }{\sigma }\\=\frac{19-15}{4}\\=1[/tex]
So,
[tex]P\left ( X\leq 19 \right )\\=P\left ( Z\leq 1 \right )\\=0.8413[/tex]