Solution: We are given that females have pulse rates that are normally distributed with a [tex]\mu=76,\sigma=12.5[/tex]
We have to find [tex]P(\bar{x}<83)[/tex]
First we need to determine the z score corresponding to [tex]\bar{x}=83[/tex]
We know that:
[tex]z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}} }[/tex]
[tex]=\frac{83-76}{\frac{12.5}{\sqrt{4}}}[/tex]
[tex]=1.12[/tex]
Now, we have to find [tex]P(z<1.12)[/tex]
Using the standard normal table, we have:
[tex]P(z<1.12)=0.8686[/tex]
Therefore, if 4 adult females are randomly selected the probability that they have pulse rates with a mean less than 83 beats per minute is 0.8686
