Question

1. find each of the following probabilities for a normal distribution.

a. (p(-1.80 < z < 0.20))
b. (p(-0.40 < z < 1.40))
c. (p(0.25 < z < 1.25))
d. (p(-0.90 < z < -0.60))

1. find the z - score location of a vertical line that separates a normal distribution as described in each of the following.

a. 5% in the tail on the right
b. 20% in the tail on the left

Explanation:

Step1: Recall the properties of the standard - normal distribution

We use the standard - normal table (z - table) which gives the cumulative probability $P(Z\leq z)$. For $P(a < Z < b)=P(Z < b)-P(Z < a)$.

Step2: Solve part a

For $P(-1.80 < Z < 0.20)$, from the z - table, $P(Z < 0.20)=0.5793$ and $P(Z < - 1.80)=0.0359$. Then $P(-1.80 < Z < 0.20)=0.5793 - 0.0359 = 0.5434$.

Step3: Solve part b

For $P(-0.40 < Z < 1.40)$, from the z - table, $P(Z < 1.40)=0.9192$ and $P(Z < - 0.40)=0.3446$. Then $P(-0.40 < Z < 1.40)=0.9192 - 0.3446 = 0.5746$.

Step4: Solve part c

For $P(0.25 < Z < 1.25)$, from the z - table, $P(Z < 1.25)=0.8944$ and $P(Z < 0.25)=0.5987$. Then $P(0.25 < Z < 1.25)=0.8944 - 0.5987 = 0.2957$.

Step5: Solve part d

For $P(-0.90 < Z < -0.60)$, from the z - table, $P(Z < - 0.60)=0.2743$ and $P(Z < - 0.90)=0.1841$. Then $P(-0.90 < Z < -0.60)=0.2743 - 0.1841 = 0.0902$.

For question 11:

Step1: Recall the properties of the z - table for tail probabilities

If the tail probability is $p$, and we want to find the z - score $z$. For a right - tail probability $p$, we find $1 - p$ in the body of the z - table. For a left - tail probability $p$, we directly find $p$ in the body of the z - table.

Step2: Solve part a

If there is 5% ($p = 0.05$) in the right - tail, then the cumulative probability to the left of the z - score is $1 - 0.05=0.95$. Looking up 0.95 in the z - table, the z - score is approximately $z = 1.645$.

Step3: Solve part b

If there is 20% ($p = 0.20$) in the left - tail, looking up 0.20 in the z - table, the z - score is approximately $z=-0.84$.

Answer:

a. $P(-1.80 < Z < 0.20)=0.5434$
b. $P(-0.40 < Z < 1.40)=0.5746$
c. $P(0.25 < Z < 1.25)=0.2957$
d. $P(-0.90 < Z < -0.60)=0.0902$
11.a. $z = 1.645$
11.b. $z=-0.84$