4. high school students live an average of 5.6 miles from school with a standard deviation of 0.9 miles.\na…

Explanation:

Step1: Standardize the values

We use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 5.6$ (mean) and $\sigma=0.9$ (standard deviation).

Step2: Solve part a

For $x = 4$, $z_1=\frac{4 - 5.6}{0.9}=\frac{-1.6}{0.9}\approx - 1.78$. For $x = 6$, $z_2=\frac{6 - 5.6}{0.9}=\frac{0.4}{0.9}\approx0.44$. Using the standard normal table, $P(-1.78<Z<0.44)=\Phi(0.44)-\Phi(-1.78)$. Since $\Phi(-z)=1 - \Phi(z)$, $\Phi(-1.78)=1-\Phi(1.78)$. From the standard - normal table, $\Phi(0.44) = 0.6700$, $\Phi(1.78)=0.9625$, so $P(-1.78<Z<0.44)=0.6700-(1 - 0.9625)=0.6325 = 63.25%$.

Step3: Solve part b

For $x = 7$, $z=\frac{7 - 5.6}{0.9}=\frac{1.4}{0.9}\approx1.56$. Using the standard normal table, $P(Z\leq1.56)=\Phi(1.56)=0.9406 = 94.06%$.

Step4: Solve part c

For $x = 6$, $z=\frac{6 - 5.6}{0.9}\approx0.44$. Using the standard normal table, $P(Z\geq0.44)=1 - P(Z < 0.44)=1 - 0.6700 = 0.3300=33.00%$.

Step5: Solve part d

We want to find $x$ when $P(Z<z)=0.30$. Looking up in the standard - normal table, the $z$ - value corresponding to a probability of $0.30$ is approximately $z=-0.52$. Then, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we have $-0.52=\frac{x - 5.6}{0.9}$. Solving for $x$ gives $x=5.6+(-0.52)\times0.9=5.6 - 0.468 = 5.132$ miles.

Step6: Solve part e

The $z$ - value corresponding to the 75th percentile is $z = 0.67$ (from the standard - normal table). Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we have $0.67=\frac{x - 5.6}{0.9}$. Solving for $x$ gives $x=5.6+0.67\times0.9=5.6 + 0.603=6.203$ miles.

Step7: Solve part f

If only 10% of students live farther than Kaliq, then $P(Z>z)=0.10$, so $P(Z\leq z)=0.90$. Looking up in the standard - normal table, $z = 1.28$. Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we have $1.28=\frac{x - 5.6}{0.9}$. Solving for $x$ gives $x=5.6+1.28\times0.9=5.6 + 1.152 = 6.752$ miles.

Step8: Solve part g

Kaliq lives approximately 6.752 miles from school. Gabby lives 6.5 miles from school. Since $6.5<6.752$, Gabby's statement is incorrect.

Answer:

a. $63.25%$ b. $94.06%$ c. $33.00%$ d. $5.132$ e. $6.203$ f. $6.752$ g. Incorrect, because Kaliq lives approximately 6.752 miles and Gabby lives 6.5 miles.