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.
0.005
Multiplying by ten moves the decimal point over one.
Answer:
9.62162162162
Step-by-step explanation:
Use a calculator dude
Answer:
Option C
Step-by-step explanation:
Answer:
1/2
Step-by-step explanation:
Given data
Total melon= 1 1/2
In simple fraction= 3/2
DIvide 3/2 by 2
= 3/2/3
=3/2 * 1/3
=3/6
= 1/2
Hence each person will get 1/2 of the whole water melon