Answer to Question #261476 in Statistics and Probability for Joyce

Question

The weight of baby kangaroos are known to have a mean of 125 pounds and a standard deviation of 15 pounds. If we obtained a random sample of 40 baby kangaroos what is the probability that the sample mean will be between 120 and 130 pounds and what are the confidence intervals having the same number of samples with a 95% confidence level?

Answer 1

Let $X=$ the sample mean: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=125, \sigma=15, n=40.$

i)

$$P(120<X<130)=P(X<130)-P(X\leq 120)$$

$$=P(Z<\dfrac{130-125}{15/\sqrt{40}})-P(Z\leq \dfrac{120-125}{15/\sqrt{40}})$$

$$\approx P(Z<2.108185)-P(Z\leq-2.108185)$$

$$\approx0.9824925-0.0175075$$

$$\approx0.964985$$

ii) The critical value for $\alpha = 0.05$ is$z_c = z_{1-\alpha/2} = 1.96.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-z_c\times \dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times \dfrac{\sigma}{\sqrt{n}})$$

$$=(125-1.96\times \dfrac{15}{\sqrt{40}}, 125+1.96\times \dfrac{15}{\sqrt{40}})$$

$$=(120.352, 129.648)$$

Therefore, based on the data provided, the 95% confidence interval for the population mean is $120.352 < \mu < 129.648,$ which indicates that we are 95% confident that the true population mean $\mu$ is contained by the interval $(120.352, 129.648).$