(a) Find the probability that a single randomly selected value is less than 78.8. Round your answer to four decimal places.
(b) Find the probability that a randomly selected sample of size n = 38 has a mean less than 78.8. Round your answer to four decimal places.
Answer :
Given :
The population mean $(μ) = 68.1$
The population standard deviation $(σ) = 20.7$
Solution :
(a) The probability that a single randomly selected value is less than 78.8 :
$$P(x < 78.8) = P\left(\frac{x-\mu}{\sigma} < \frac{78.8 – 68.1}{20.7}\right)$$ $$= P(z < 0.517)$$ $$= 0.6974$$
(b) The probability that a randomly selected sample of size $n=38$ has a mean less than 78.8 :
$$P(\bar{x} < 78.8) = P\left(\frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}} < \frac{78.8 – 68.1}{\frac{20.7}{\sqrt{38}}}\right)$$ $$= P(z < 3.186)$$ $$= 0.9993$$
