Question

a normal population has mean μ=7 and standard deviation σ=7. (a) what proportion of the population is less than 19? (b) what is the probability that a randomly chosen value will be greater than 4? round the answers to four decimal places. part 1 of 2 the proportion of the population less than 19 is 0.9568. part: 1 / 2 part 2 of 2 the probability that a randomly chosen value will be greater than 4 is .

Explanation:

Step1: Calculate the z - score

For a normal distribution, the z - score is calculated using the formula $z=\frac{x - \mu}{\sigma}$. Here, $x = 4$, $\mu=7$, and $\sigma = 7$. So, $z=\frac{4 - 7}{7}=\frac{- 3}{7}\approx - 0.4286$.

Step2: Find the probability $P(X>4)$

We know that $P(X > 4)=1 - P(X\leq4)$. And $P(X\leq4)$ is the cumulative probability corresponding to the z - score $z=-0.4286$. Looking up the z - table or using a calculator, $P(Z\leq - 0.43)\approx0.3336$ (using more precise calculation for $z=-0.4286$: the cumulative distribution function for standard normal $\Phi(-0.4286)=1-\Phi(0.4286)$. $\Phi(0.4286)\approx0.6664$, so $1 - 0.6664 = 0.3336$? Wait, no, actually $P(X\leq4)=P(Z\leq\frac{4 - 7}{7})=P(Z\leq - 0.4286)$. The value of $\Phi(-0.4286)$ can be calculated as $1-\Phi(0.4286)$. $\Phi(0.4286)\approx0.6664$, so $P(Z\leq - 0.4286)=1 - 0.6664 = 0.3336$? Wait, no, let's use a more accurate method. Using the standard normal table or a calculator, for $z=-0.43$, $\Phi(-0.43)=0.3336$, for $z = - 0.42$, $\Phi(-0.42)=0.3372$. Since our $z=-0.4286$ is closer to $-0.43$, we can use linear approximation. The difference between $z=-0.42$ and $z=-0.43$ is $0.01$ in z - score, and the difference in $\Phi(z)$ is $0.3372 - 0.3336 = 0.0036$. Our z - score is $-0.4286=-0.42-0.0086$. So the decrease in $\Phi(z)$ from $z=-0.42$ is $0.0086\times\frac{0.0036}{0.01}=0.003096$. So $\Phi(-0.4286)\approx0.3372-0.003096 = 0.3341$. Then $P(X > 4)=1 - P(X\leq4)=1 - 0.3341 = 0.6659$ (more accurately, using a calculator, $P(Z > - 0.4286)=1 - \Phi(-0.4286)=\Phi(0.4286)\approx0.6664$ (because the standard normal distribution is symmetric, $P(Z > - z)=\Phi(z)$). Wait, I made a mistake earlier. The correct formula is $P(X>4)=P(Z>\frac{4 - 7}{7})=P(Z > - 0.4286)=1 - P(Z\leq - 0.4286)=\Phi(0.4286)$. $\Phi(0.4286)$: looking at the standard normal table, for $z = 0.42$, $\Phi(0.42)=0.6628$, for $z = 0.43$, $\Phi(0.43)=0.6664$. The value of $z = 0.4286$ is $0.42+0.0086$. The difference between $z = 0.42$ and $z = 0.43$ is $0.01$ in z - score, and the difference in $\Phi(z)$ is $0.6664 - 0.6628 = 0.0036$. So the increase in $\Phi(z)$ from $z = 0.42$ is $0.0086\times\frac{0.0036}{0.01}=0.003096$. So $\Phi(0.4286)\approx0.6628 + 0.003096=0.6659$. So $P(X > 4)\approx0.6659$.

Answer:

0.6659