a. If we select a random sample of 48 households, what is the standard error of the mean?
b. What is the expected shape of the distribution of the sample mean?
c. What is the likelihood of selecting a sample with a mean more than $120,000?
d. What is the likelihood of selecting a sample with a mean of more than $106,000?
e. Find the likelihood of selecting a sample with a mean of more than 106,000 but less than 120,000.
Answer:
Given data,
The population mean, $μ=115,000$
The population standard deviation, $σ=37,000$
(a) The sample size, $n=48$
The standard error of the mean can be calculated as,
$$\therefore \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} = \frac{37{,}000}{\sqrt{48}} = 5{,}340.49 = 5{,}340$$
(b)
→ As the sample size is greater than 30, according to the Central Limit Theorem, the distribution of the sample mean can be approximated as normal.
∴ The distribution of sample mean will be normal.
(c) The likelihood of selecting a selecting a sample with a mean more than $120,000:
$$P(\overline{x} > 120{,}000) = P\left(\frac{\overline{x} – \mu}{\frac{\sigma}{\sqrt{n}}} > \frac{120{,}000 – 115{,}000}{\frac{37{,}000}{\sqrt{48}}}\right)$$ $$= P(z > 0.94)$$ $$= 1 – P(z < 0.94)$$ $$= 1 – 0.8264$$ $$= 0.1736$$
(d) The likelihood of selecting a selecting a sample with a mean more than $106,000:
$$P(\overline{x} > 106{,}000) = P\left(\frac{\overline{x} – \mu}{\frac{\sigma}{\sqrt{n}}} > \frac{106{,}000 – 115{,}000}{\frac{37{,}000}{\sqrt{48}}}\right)$$ $$= P(z > -1.69)$$ $$= 1 – P(z < -1.69)$$ $$= 1 – 0.0455$$ $$= 0.9545$$
(e) The likelihood of selecting a sample with a mean of more than 106,000 but less than 120,000:
$$P(106{,}000 < \overline{x} < 120{,}000) = P\left(\frac{106{,}000 – 115{,}000}{\frac{37{,}000}{\sqrt{48}}} < \frac{\overline{x} – \mu}{\frac{\sigma}{\sqrt{n}}} < \frac{120{,}000 – 115{,}000}{\frac{37{,}000}{\sqrt{48}}}\right)$$ $$= P(-1.69 < z < 0.94)$$ $$= P(z < 0.94) – P(z < -1.69)$$ $$= 0.8264 – 0.0455$$ $$= 0.7809$$
