Z is a standard normal random variable. The P(1.05 < Z < 2.13) equals 0.8365 0.1
ID: 3271941 • Letter: Z
Question
Z is a standard normal random variable. The P(1.05 < Z < 2.13) equals
0.8365
0.1303
0.4834
0.3531
Z is a standard normal random variable. The P(Z > 2.11) equals
0.4821
0.9821
0.5000
0.0174
Z is a standard normal random variable. The P(1.5 < Z < 1.09) equals
0.4322
0.3621
0.7953
0.0711
Given that Z is a standard normal random variable, what is the probability that -2.51 Z -1.53?
0.4950
0.4370
0.0570
0.9310
Given that Z is a standard normal random variable, what is the probability that Z -2.12?
0.4830
0.9830
0.0170
0.9660
Given that Z is a standard normal random variable, what is the probability that -2.08 Z 1.46?
0.9091
0.4812
0.4279
0.01876
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
Refer to Exhibit 6-2. The probability of a player weighing more than 241.25 pounds is
0.4505
0.0495
0.9505
0.9010
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
Refer to Exhibit 6-2. The probability of a player weighing less than 250 pounds is
0.4772
0.9772
0.0528
0.5000
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
Refer to Exhibit 6-2. What percent of players weigh between 180 and 220 pounds?
28.81%
0.5762%
0.281%
57.62%
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
Refer to Exhibit 6-2. What is the minimum weight of the middle 95% of the players?
196
151
249
190
The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000.
Refer to Exhibit 6-6. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?
0.4772
0.9772
0.0228
0.5000
The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000.
Refer to Exhibit 6-6. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $47,500?
0.4332
0.9332
0.0668
0.5000
The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000.
Refer to Exhibit 6-6. What percentage of MBA's will have starting salaries of $34,000 to $46,000?
38.49%
38.59%
50%
76.98%
0.8365
0.1303
0.4834
0.3531
Explanation / Answer
(a) P(1.05 < Z < 2.13):
For Z value from mid value to 1.05 on RHS of mid value, area from Table = 0.3531
For Z value from mid value to 2.13 on RHS of mid valye, area from Table = 0.4834
So, P(1.05 < Z < 2.13) = 0.4834 - 0.3531 = 0.1303
So, correct option:
B 0.1303
(b) P(Z>2.11):
For Z value from mid value to 2.11 on RHS of mid value, area from Table = 0.4826
So, P(Z>2.11) = 0.5 - 0.4826 = 0.0174
So, correct option:
D 0.0174
(c) P(-1.5 < Z < 1.09):
For Z value from mid value to - 1.5 on LHS of mid value, area from Table = 0.4332
For Z value from mid value to 1.09 on RHS of mid value, area from Table = 0.3621
So, P(-1.5 < Z < 1.09) = 0.4332 + 0,3621 = 0.7953
So, correct option:
C 0.7953
(d) P(-2.51 < Z < - 1.53):
For Z value from mid value to - 2.51 on LHS of mid value, area from Table = 0.4940
For Z value from mid value to - 1.53 on LHS of mid value, area from Table = 0.4370
So, P(-2.51 < Z < - 1.53) = 0.4940 - 0.4370 = 0.0570
So, correct option:
C 0.0570
ONLY FIRST 4 QUESTIONS ANSWERED AS PER DIRECTIONS FOR ANSWERING.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.