1. Find the probability P(Z < 0.27) using the standard normaldistribution. A) 0.

Question

1. Find the probability P(Z < 0.27) using the standard normaldistribution.
A) 0.7300 B) 0.3936 C) 0.6064 D) 0.1064

2. Find the probability P(Z < –0.46) using the standardnormal distribution.
A) 0.6772 B) 0.3228 C) 0.8228 D) 0.5400

3. Find the probability P(Z > –0.74) using the standardnormal distribution.
A) 0.2296 B) 0.7704 C) 0.7296 D) 0.2600

4. Find the probability P( -2.05 <Z< 2.05).
A) 0.4938 B) 0.4798 C) 0.9596 D) 0.9876

5. Find the probability P(Z<0 or Z> 1.67).
A)
0.45
C)
0.35
B)
0.55
D)
0.65

6. Find the probability P(0.36 < z < 1.43) using the standardnormal distribution.
A) 0.2831 B) 0.7169 C) 0.7831 D) 0.5600

7. The mean gas mileage of a certain model car is 28 miles pergallon. If the gas mileages are normally distributed with astandard deviation of 1.7, find the probability that a car has agas mileage of between 27.8 and 28.3 miles per gallon.
A) 0.15 B) 0.17 C) 0.12 D) 0.23

8. A recent study found that the mean life expectancy of a personliving in Africa is 53 years with a standard deviation of 7.5years. If a person in Africa is selected at random, what is theprobability that the person will die before the age of 65?
A) 94.52% B) 82.89% C) 94.95% D) 88.49%

9. At a large department store, the average number of years ofemployment for a cashier is 5.7 with a standard deviation of 1.8years. If an employee is picked at random, what is the probabilitythat the employee has worked at the store for over 10 years?
A) 49.16% B) 99.16% C) 0.84% D) 0.54%

Explanation / Answer

1) C) 0.6064 2) B) 0.3228 3) B) 0.7704 4) 1-2(0.0202) = 0.9596 Answer: C 5) P(Z<0 or Z> 1.67) = 0.50 + 0.0475 = 0.5475 Answer: B 6) P(0.36 < z < 1.43) = 0.3594 - 0.0764 = 0.283 Answer: A

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