a. The area to the left of z is 0.9750
Answer: We can use the excel formula to find the value of z corresponding to an area 0.9750. The excel formula is:
Therefore, the area to the left of 1.96 is 0.9750
b. The area between 0 and z is .4750.
Answer: Here we have to find the value of z, knowing that the area between 0 and z is 0.4750.
We know the area less than z = 0.5 + 0.4750 = 0.9750. Now we can use the below excel formula to find the value of z.
Therefore, the area between 0 and 1.96 is 0.4750
c. The area to the left of z is .7291.
Answer: We can use the excel formula to find the value of z corresponding to an area 0.7291. The excel formula is:
Therefore, the area to the left of 0.61 is 0.9750
d. The area to the right of z is .1314.
Answer: Please note the excel gives us only the z-values for the areas to the left of z. Now if the area to the right of z is 0.1314, it means the area to the left of z would be 1 - 0.1314 = 0.8686. Now we can use the excel formula, as:
Therefore, the area to the right of 1.12 is .1314.
e. The area to the left of z is .6700.
Answer: We can use the excel formula to find the value of z corresponding to an area 0.6700. The excel formula is:
Therefore, the area to the left of 0.44 is 0.6700
f. The area to the right of z is .3300.
Answer: Please note the excel gives us only the z-values for the areas to the left of z. Now if the area to the right of z is 0.3300, it means the area to the left of z would be 1 - 0.3300= 0.6700. Now we can use the excel formula, as:
Therefore, the area to the right of 0.44 is 0.3300
please solve this problem using Excel step by step need to understand how thr problem is...
4) Given that z is a standard normal random variable, find z for each situation (using excel): The area to the left of z is .9750. The area between 0 and z is .4750. The area to the left of z is .7291. The area to the right of z is .1314. The area to the left of z is .6700. The area to the right of z is .3300.
4) Given that z is a standard normal random variable, find z for each situation The area to the left of z is .9750. The area between 0 and z is .4750. The area to the left of z is .7291. The area to the right of z is .1314. The area to the left of z is .6700. The area to the right of z is .3300.
You may need to use the appropriate appendix table to answer this question. Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.) (a) The area to the left of z is 0.1841 (b) The area between -z and z is 0.9398. (c) The area between -z and z is 0.2282 (d) The area to the left of z is 0.9951. (e) The area to the right of z...
2. Given that z is a standard normal random variable, compute the following probabilities. P(-1 ≤ z ≤ 0) (Round to four decimal places) Answer P(-1.5 ≤ z ≤ 0) (Round to four decimal places) Answer P(-2 < z < 0) (Round to four decimal places) Answer P(-2.5 < z < 0) (Round to four decimal places) Answer P(-3 ≤ z ≤ 0) (Round to four decimal places) 3. Given that z is a standard normal random variable, compute the...
please answer the question in Excel step by step 2) Given that z is a standard normal random variable, compute the following probabilities. a. P(z ≤ -1.09) b. P(z ≤ -1.5) c. P(z ≥ 1.3) d. P(z ≥ 2.54) e. P(-0.71 < z ≤ 2.54)
Given that z is a standard normal random variable, find z for each situation (to 2 decimals) a. The area to the left of z is 0.2119 b. The area between -z and z is 0.903. c. The area between -z and z is 0.2052 d. The area to the left of z is 0.9951 e. The area to the right of z is 0.695
Given that is a standard normal random variable, find for each situation (to 2 or 3 decimals). a. The area to the left of z is 0.8 b. The area to the left of z is 0.981 c. The area to the right of z is 0.6045 d. The area between -z and z is 0.82 (Hint: Enter the positive z-value)
eBook Video Given that z is a standard normal random variable, find z for each situation (to 2 decimals). Enter negative values as negative numbers. a. The area to the left of z is 0.2119. 1.66 b. The area between – 2 and z is 0.9030. 1.66 c. The area between – z and z is 0.2052 . 1 .26 d. The area to the left of z is 0.9948 . e. The area to the right of z is...
Please solve #22 with step by step. I need to understand clearly from your solution. also, please use the Hint. When you provide the solution, please write it clearly to be recognizable. 22. Find Z[GL2(R)]. 6. (Center of the matrix group.) Solve Problem 22, Section 2.3, asking you to find the center of the group GL2(R). Hint: Let A= ( 2) be in the center, and write down explicitly the condition that AB = BA for all matrices B= (%)...
eBook Video Given that z is a standard normal random variable, find z for each situation (to 2 decimals). a. The area to the left of z is 0.209. (Enter negative value as negative number.) -0.81 b. The area between – z and z is 0.905. 1.1553 c. The area between – z and z is 0.2128 . d. The area to the left of z is 0.9951 . -2.58 e. The area to the right of z is 0.6915....