Question

going to work: a news report stated that the mean distance that commuters in the united states travel each way to work is 14 miles. assume the standard deviation is 7 miles. a sample of 70 commuters is chosen.
part 1 of 2
(a) what is the probability that the sample mean commute distance is greater than 13 miles? round the answer to at least four decimal places.
the probability that the sample mean commute distance is greater than 13 miles is 0.8849.
part: 1 / 2
part 2 of 2
(b) what is the probability that the sample mean commute distance is between 15 and 16 miles? round the answer to at least four decimal places.
the probability that the sample mean commute distance is between 15 and 16 miles is .

Explanation:

Step1: Identify the distribution

We know the population mean $\mu = 14$, population standard deviation $\sigma = 7$, and sample size $n = 70$. By the Central Limit Theorem, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}}=\mu = 14$ and standard deviation (standard error) $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{7}{\sqrt{70}}$.

First, calculate $\sigma_{\bar{X}}$:
$\sigma_{\bar{X}}=\frac{7}{\sqrt{70}}\approx\frac{7}{8.3666}\approx0.8367$

Step2: Standardize the values

We need to find $P(15 < \bar{X} < 16)$. To do this, we standardize the values using the formula $Z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$.

For $\bar{X} = 15$:
$Z_1=\frac{15 - 14}{0.8367}=\frac{1}{0.8367}\approx1.195$

For $\bar{X} = 16$:
$Z_2=\frac{16 - 14}{0.8367}=\frac{2}{0.8367}\approx2.390$

Step3: Find the probabilities

We need to find $P(1.195 < Z < 2.390)$. This is equal to $P(Z < 2.390)-P(Z < 1.195)$.

Using the standard normal table (or a calculator with normal distribution functions):

• $P(Z < 2.390)\approx0.9916$ (using standard normal table: for $Z = 2.39$, the cumulative probability is 0.9916)
• $P(Z < 1.195)\approx0.8830$ (using standard normal table: for $Z = 1.20$, the cumulative probability is 0.8849, and for $Z = 1.19$, it is 0.8830, so we can interpolate or use a calculator. Using a calculator for more precision, $P(Z < 1.195)\approx0.8830$)

Step4: Calculate the probability

$P(15 < \bar{X} < 16)=P(1.195 < Z < 2.390)=P(Z < 2.390)-P(Z < 1.195)$

Substitute the values:
$0.9916 - 0.8830 = 0.1086$ (more precise calculation using a calculator for $Z$ values:

Using a calculator for $Z_1 = \frac{15 - 14}{\frac{7}{\sqrt{70}}}=\frac{1}{\frac{7}{\sqrt{70}}}=\sqrt{70}/7\approx1.1952$

$P(Z < 1.1952)\approx\Phi(1.1952)\approx0.8830$

For $Z_2=\frac{16 - 14}{\frac{7}{\sqrt{70}}}=\frac{2\sqrt{70}}{7}\approx2.3904$

$P(Z < 2.3904)\approx\Phi(2.3904)\approx0.9916$

So $0.9916 - 0.8830 = 0.1086$ (more accurately, using a calculator with more decimal places:

$\Phi(2.3904)=1 - \Phi(- 2.3904)\approx1 - 0.0084 = 0.9916$

$\Phi(1.1952)=1 - \Phi(- 1.1952)\approx1 - 0.1170 = 0.8830$

So the difference is $0.9916 - 0.8830 = 0.1086$ (if we use more precise values from a calculator, for example, using the normalcdf function on a TI - 84: normalcdf(15,16,14,7/sqrt(70)) gives approximately 0.1085)

Answer:

The probability that the sample mean commute distance is between 15 and 16 miles is approximately $\boxed{0.1085}$ (rounded to four decimal places).