Question

In 2018 in an attempt to improve the reputation of the Democratic People’s Republic of Korea (DPRK) lottery tickets were sold to people around the world. The grand prize of this lottery was a weekend with Kim Jung Un. During anevening with Kim Jung Un the lottery winner was offered a meal made from one of the lobsters in Kim Jung Un’s private lobster aquarium.(Which by the way are all Maine lobsters!) The average weight of the lobsters was 22 ounces and the standard deviation was 0.67 ounces. When a random lobster wastaken from Kim Jung Un’s aquarium what was the probability it weighed more than 23.75 ounces?
a.) 0.0154 b.) 0.9955 c.) 0.9846 d.) 0.0045 e.)None of these

In lieu of using a single resistor three resistors are wired in series. The three resistors are identical. The resistance o f each is normally distributed with a mean of 6 ohms and a standard deviation of 0.3 ohms. The probability the combined resistance will exceed 19 ohm's is 0.0274. How precise (i.e. what is the required value of the standard deviation) would the manufacturing process have to be make the probability less than 0.0055 that t he combined resistance of the circuit would exceed 19 ohms?
a.) 0.180 ohms b.) 0.220 ohms c.) 0.227 ohms d.) 0.229 ohms e.) None of these

An experiment has two possible outcomes: the first occurs with probability p ; the second with probability p^2 . What is p?
a.) 0.3820 b.) 0.5000    c.) 0.2500     d.)   0.6180 e.) None of the above

Of all 3–to–5year old children, 56% are enrolled in school. If a sample of 500 such children is randomly selected, find the probability that at least 250 will be enrolled in school.Hint: Use De Moivre–Laplace.
a.) 0.9970 b.) 0.0035 c.) 0.9965 d.) 0.0030 e.) None of the above

Answer 1

1)

µ =    22                  
σ =    0.67                  
                      
P ( X ≥   23.75   ) = P( (X-µ)/σ ≥ (23.75-22) / 0.67)              
= P(Z ≥   2.61   ) = P( Z <   -2.612   ) =    0.00450   (answer)

3)

Sample size , n =    500          
Probability of an event of interest, p =   0.56          
right tailed              
X ≥   250          
              
Mean = np =    280          
std dev ,σ=√np(1-p)=   11.0995          
              
P(X ≥   250   ) = P(Xnormal ≥   249.5   )
              
Z=(Xnormal - µ ) / σ =     (249.5-280)/11.0995)=   -2.748      
              
=P(Z ≥   -2.748   ) =    0.9970