the time needed for passengers to board the twisting thunder roller coaster is skewed right with a mean of…

Explanation:

Step1: Define variables

Let $\bar{X}_1$ be the sample - mean time for Twisting Thunder and $\bar{X}_2$ be the sample - mean time for Spiral Wonder. We know that $\mu_1 = 49$, $\sigma_1 = 7.1$, $n_1 = 32$, $\mu_2 = 44.8$, $\sigma_2 = 3.7$, $n_2 = 36$. The mean of the sampling distribution of $\bar{X}_1-\bar{X}_2$ is $\mu=\mu_1 - \mu_2=49 - 44.8 = 4.2$. The standard deviation of the sampling distribution of $\bar{X}_1-\bar{X}_2$ is $\sigma=\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{\frac{7.1^{2}}{32}+\frac{3.7^{2}}{36}}$. First, calculate $\frac{7.1^{2}}{32}+\frac{3.7^{2}}{36}$: $\frac{7.1^{2}}{32}=\frac{50.41}{32}\approx1.5753$ $\frac{3.7^{2}}{36}=\frac{13.69}{36}\approx0.3803$ $\frac{7.1^{2}}{32}+\frac{3.7^{2}}{36}\approx1.5753 + 0.3803=1.9556$ $\sigma=\sqrt{1.9556}\approx1.4$

Step2: Standardize the value

We want to find $P(\bar{X}_1<\bar{X}_2)$, which is equivalent to $P(\bar{X}_1 - \bar{X}_2<0)$. Let $Z=\frac{(\bar{X}_1 - \bar{X}_2)-\mu}{\sigma}$. Substituting the values, we get $Z=\frac{0 - 4.2}{1.4}=- 3$.

Step3: Find the probability

We know that $P(Z < - 3)$ can be found using the standard - normal distribution table. From the standard - normal table, $P(Z < - 3)=0.0013$.

Answer:

0.0013