Find the probability that a sample of size n=109 is randomly selected with a mean between 76.1 and 79.8.
P(76.1 < \bar{x} < 79.8) =
Enter your answers as numbers accurate to 4 decimal places.
Answer:
Given:
The population mean $(𝜇) = 79.4$
The population standard deviation $(𝜎) = 13.2$
The sample size $(n) = 109$
→ The probability that a sample of size n=109 is randomly selected with a mean between 76.1 and 79.8:
$$P(76.1 < \overline{x} < 79.8) = P\left(\frac{76.1 – 79.4}{\frac{13.2}{\sqrt{109}}} < \frac{\overline{x} – \mu}{\frac{\sigma}{\sqrt{n}}} < \frac{79.8 – 79.4}{\frac{13.2}{\sqrt{109}}}\right) $$ $$= P(-2.61 < z < 0.316) $$ $$= P(z < 0.316) – P(z < -2.61) $$ $$= 0.624 – 0.0045 $$ $$ = 0.6195$$
