Using the standard normal distribution find the following. а. Р(z $2.39) e. P(-2.35< z<1.87) b. Р(z21.28) f. P(1.28

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Please do only sub-part "f, g, h" for Q7 and also do only sub-part "d" for Q9.

Don't use excel formula to solve the question.

### Using the Standard Normal Distribution

7. Find the following probabilities using the standard normal distribution.

a. \( P(z \leq 2.39) \)

b. \( P(z \geq 1.28) \)

c. \( P(z \leq -2.87) \)

d. \( P(z \geq -3.21) \)

e. \( P(-2.35 \leq z \leq 1.87) \)

f. \( P(1.28 \leq z \leq 2.05) \)

g. Find \( z^* \) so that \( P(-z^* \leq z \leq z^*) = 0.98 \)

h. Find \( z^* \) so that \( P(-z^* \leq z \leq z^*) = 0.99 \)
Transcribed Image Text:### Using the Standard Normal Distribution 7. Find the following probabilities using the standard normal distribution. a. \( P(z \leq 2.39) \) b. \( P(z \geq 1.28) \) c. \( P(z \leq -2.87) \) d. \( P(z \geq -3.21) \) e. \( P(-2.35 \leq z \leq 1.87) \) f. \( P(1.28 \leq z \leq 2.05) \) g. Find \( z^* \) so that \( P(-z^* \leq z \leq z^*) = 0.98 \) h. Find \( z^* \) so that \( P(-z^* \leq z \leq z^*) = 0.99 \)
9. Weights of walleye in a lake are normally distributed with a mean of 4.8 pounds and a standard deviation of 1.3 pounds.

a. What percentage of walleye weighs below 3 pounds?

b. What percentage of walleye weighs between 2 and 7 pounds?

c. The DNR decided to catch the bottom 10% of the walleye and release them to another lake. What weight must a fish have in order to be relocated to the other lake?

d. Find the 95th percentile of the weights of walleye.
Transcribed Image Text:9. Weights of walleye in a lake are normally distributed with a mean of 4.8 pounds and a standard deviation of 1.3 pounds. a. What percentage of walleye weighs below 3 pounds? b. What percentage of walleye weighs between 2 and 7 pounds? c. The DNR decided to catch the bottom 10% of the walleye and release them to another lake. What weight must a fish have in order to be relocated to the other lake? d. Find the 95th percentile of the weights of walleye.
Expert Solution
Step 1

Hey, since there are multiple subparts posted, we will answer first three question. If you want any specific question to be answered then please submit that question only or specify the question number in your message.

(f)

The value of P1.28z2.05 is 0.0801, which is obtained below:

P1.28z2.05=Pz1.28-Pz2.05   Use Standard Normal Table=0.97982-0.89973=0.0801

(g)

The value of P-z*zz*=0.98 is -2.326 and 2.326, which is obtained below:

The left of z* and the right of z* is obtained Excel function “=NORMSINV(Probability)”.

Output obtained from Excel is given below:

Statistics homework question answer, step 1, image 1

Statistics homework question answer, step 1, image 2

From the output, the left of z* and the right of z* is -2.326 and 2.326.

 

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