这是一份GLA格拉斯哥大学STATS5022_1/STATS4047_1作业代写的成功案例

Here the population mean and the standard deviation are $\mu=110$ and $\sigma=10$, respectively. The sample size $n=75$ is large, so the central limit theorem ensures that the distribution of $\bar{X}$ is approximately normal with
Mean of $\bar{X}=110$
Standard deviation of $\bar{X}=\frac{\sigma}{\sqrt{n}}=\frac{10}{\sqrt{75}}=1.155$
To find $P[109<\bar{X}<112]$ we convert to the standardized variable
$$
Z=\frac{\bar{X}-110}{1.155}
$$
and calculate the z-values
$$
\frac{109-110}{1.155}=-.866, \quad \frac{112-110}{1.155}=1.732
$$

STATS5022_1/STATS4047_1 COURSE NOTES ：

$$
\bar{x}=\$ 227 \quad \text { and } \quad s=\$ 15
$$
With $1-\alpha=.90$ we have $\alpha / 2=.05$, and $z_{\alpha / 2}=1.645$
$$
1.645 \frac{s}{\sqrt{n}}=\frac{1.645 \times 15}{\sqrt{75}}=2.85
$$
Hence, a $90 \%$ confidence interval for the population mean $\mu$ is