There are 90 people in the restaurant The probability of som

There are 90 people in the restaurant. The probability of someone ordering a drink with the food is 60%. Use Normal approximation of Binomial Distribution to answer the following 6 questions.

1. What is the mean of the Normal distribution?

2. What is the standard deviation of the normal distribution?

3. What is the probability that exactly 50 people will order a drink?

4. What the probability that more than 50 people will order a drink?

5. What is the probability that less than 50 people will order a drink?

6. What is the probability that between 52 or more and 56 or less people will order a drink?

      -       1.       2.       3.       4.       5.       6.      

5.92%

      -       1.       2.       3.       4.       5.       6.      

77.41%

      -       1.       2.       3.       4.       5.       6.      

40.91%

      -       1.       2.       3.       4.       5.       6.      

16.64%

      -       1.       2.       3.       4.       5.       6.      

54

      -       1.       2.       3.       4.       5.       6.      

4.65

Solution

Normal Distribution
a)
Mean ( np ) =54
b)
Standard Deviation ( npq )= 90*0.6*0.4 = 4.6476
Normal Distribution = Z= X- u / sd                   
d)
P(X > 50) = (50-54)/4.6476
= -4/4.6476 = -0.8607
= P ( Z >-0.861) From Standard Normal Table
= 0.8053
e)
P(X < 50) = (50-54)/4.6476
= -4/4.6476= -0.8607
= P ( Z <-0.8607) From Standard NOrmal Table
= 0.1947                  
f)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 52) = (52-54)/4.6476
= -2/4.6476 = -0.4303
= P ( Z <-0.4303) From Standard Normal Table
= 0.33348
P(X < 56) = (56-54)/4.6476
= 2/4.6476 = 0.4303
= P ( Z <0.4303) From Standard Normal Table
= 0.66652
P(52 < X < 56) = 0.66652-0.33348 = 0.333