***solving only the wrong parts***
PART 1
First divide 0.2052/2 = 0.1026
Now look for the number 0.1026 in the Z table. We will find it in row 0.2 and column 0.06. So, the answer for z is z = 0.2 + 0.06 = 0.26
PART 2
Given that z is a standard normal random variable, find z for each situation (to 2...
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
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...
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per night (to 4 decimals)? b. What is the probability that a hotel room costs less than $140 per night (to 4 decimals)? c. What is...
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $203 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $56. Use Table 1 in Appendix B. a. What is the probability that a hotel room costs $224 or more per night (to 4 decimals)? b. What is the probability that a hotel room costs less than $141 per night...
New York City is the most expensive city in the United States for lodging. The mean hotel room rate is $203 per night (USA Today, April 30, 2012). Assume that room rates are normally distributed with a standard deviation of $56. Use Table 1 in Appendix B. a. What is the probability that a hotel room costs $226 or more per night (to 4 decimals)? .3406 b. What is the probability that a hotel room costs less than $143 per...
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....
Given that z is a standard normal random variable, find z for each situation (to 2 decimals) a. The area to the right of z is 0.03. b. The area to the right of z is 0.045 c. The area to the right of z is 0.05 d. The area to the right of z is 0.1
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)
Check My Worl eBook New York City is the most ex standard deviation of $55. Use Table 1 in Appendix B in the United States for lodging. The mean hotel room rate is $204 per night (USA Today, April 30, 2012). Assume that room rates are normally distribute What is the probability that a hotel room costs $225 or more per night (to 4 decimals)? 0.3513 b. What is the probability that a hotel room costs less than $140 per...
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.