Question

bell ringer - oct 14
use standard normal distribution to find the areas under the bell curve.

1. p(z>-1.289)
2. p(z<0.6491)
3. with z is between -1.82 and 2.098
4. to the right of z = 2.01
5. to the left of z=-1.5309
6. p(0.5892<z<2.3518)

Explanation:

Step1: Recall standard - normal table property

The total area under the standard - normal curve is 1, and the standard - normal table gives $P(Z\leq z)$.

Step2: Solve $P(z > - 1.289)$

We know that $P(Z > z)=1 - P(Z\leq z)$. So $P(Z > - 1.289)=1 - P(Z\leq - 1.289)$. Looking up $P(Z\leq - 1.289)$ in the standard - normal table, we get approximately $0.0985$. Then $P(Z > - 1.289)=1 - 0.0985 = 0.9015$.

Step3: Solve $P(z < 0.6491)$

We directly look up $z = 0.6491$ in the standard - normal table. $P(Z < 0.6491)\approx0.7422$.

Step4: Solve $P(-1.82

$P(-1.82 < Z < 2.098)=P(Z < 2.098)-P(Z < - 1.82)$. Looking up in the table, $P(Z < 2.098)\approx0.9817$ and $P(Z < - 1.82)\approx0.0344$. So $P(-1.82 < Z < 2.098)=0.9817-0.0344 = 0.9473$.

Step5: Solve the area to the right of $z = 2.01$

$P(Z>2.01)=1 - P(Z\leq2.01)$. Looking up $P(Z\leq2.01)$ in the table, we get approximately $0.9778$. Then $P(Z > 2.01)=1 - 0.9778 = 0.0222$.

Step6: Solve the area to the left of $z=-1.5309$

Looking up $z=-1.5309$ in the standard - normal table, $P(Z < - 1.5309)\approx0.0630$.

Step7: Solve $P(0.5892

$P(0.5892 < Z < 2.3518)=P(Z < 2.3518)-P(Z < 0.5892)$. Looking up in the table, $P(Z < 2.3518)\approx0.9906$ and $P(Z < 0.5892)\approx0.7224$. So $P(0.5892 < Z < 2.3518)=0.9906 - 0.7224=0.2682$.

Answer:

1. $0.9015$
2. $0.7422$
3. $0.9473$
4. $0.0222$
5. $0.0630$
6. $0.2682$