We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.
First of all, we will find z-score corresponding to 38 and 56.


Now we will find z-score corresponding to 56.

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is  .
.
We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.
We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

Therefore, approximately  of lightbulb replacement requests numbering between 38 and 56.
 of lightbulb replacement requests numbering between 38 and 56.
 
        
             
        
        
        
Answer:
True
Step-by-step explanation:
Simple random sampling chooses at random members of the population. This allows all members an equal probability of being selected for the sample.
 
        
             
        
        
        
You can say : 15% ⇒ ? if 100% ⇒ 32.50 $ 
so next you can say : 15/100 = ?/32.50 ⇒ ? = (15/100) × 32.50 = 15 × 32.50 ÷ 100 = 4.875 ⇒ so the answer is 4.785 $ ;D i hope this will be helpful :))
        
                    
             
        
        
        
1. is D and 2. is C and 3. is B
        
             
        
        
        
Step-by-step explanation: