Answer:
[tex]P(\bar X\:<\:51.5)=0.9641[/tex] or [tex]P(\bar X\:<\:51.5)=96.41\%[/tex]
Step-by-step explanation:
Calculate the z-score of 51.5 using the formula:
[tex]z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
We substitute [tex]\bar X=51.5,n=36,\sigma=5, \mu=50[/tex]
[tex]\implies z=\frac{51.5-50}{\frac{5}{\sqrt{36} } }[/tex]
[tex]\implies z=\frac{1.5}{\frac{5}{6} }=1.80[/tex]
From the standard normal distribution table, the area to the left of 1.80 is 0.9641
Therefore [tex]P(\bar X\:<\:51.5)=0.9641[/tex]
