the monthly cost for cell phones for plan amazing has a mean of $39.17 with a standard deviation of $13.58…

Explanation:

Step1: Define variables

Let $\mu_1 = 39.17$, $\sigma_1=13.58$, $n_1 = 37$, $\mu_2 = 41.16$, $\sigma_2 = 7.18$, $n_2=40$.

Step2: Calculate the mean and standard deviation of the difference in sample - means

The mean of the difference in sample - means $\mu_{\bar{X}_1-\bar{X}2}=\mu_1-\mu_2=39.17 - 41.16=-1.99$. The standard deviation of the difference in sample - means $\sigma{\bar{X}_1-\bar{X}_2}=\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{\frac{13.58^{2}}{37}+\frac{7.18^{2}}{40}}=\sqrt{\frac{184.4164}{37}+\frac{51.5524}{40}}=\sqrt{4.9842 + 1.2888}=\sqrt{6.273}\approx2.5046$.

Step3: Calculate the z - score

We want to find $P(\bar{X}_1>\bar{X}_2)$, which is equivalent to $P(\bar{X}_1-\bar{X}2 > 0)$. The z - score is $z=\frac{0 - \mu{\bar{X}_1-\bar{X}2}}{\sigma{\bar{X}_1-\bar{X}_2}}=\frac{0-(-1.99)}{2.5046}=\frac{1.99}{2.5046}\approx0.7946$.

Step4: Find the probability

Using the standard normal distribution table, $P(Z > 0.7946)=1 - P(Z\leqslant0.7946)$. From the standard - normal table, $P(Z\leqslant0.7946)\approx0.7866$. So $P(Z > 0.7946)=1 - 0.7866 = 0.2134$.

Answer:

0.2134