Scores on an exam follow an approximately bell shaped distribution with a mean of 76.4 and a standard deviation of 6.1 points. approximately, what percentage of the data is between 64.2 points and 88.6 points?
a) 68%
b) 95%
c) 99.7%
d) 50%

b) 95%

Step-by-step explanation:

We have been given that scores on an approximately bell shaped distribution with a mean of 76.4 and a standard deviation of 6.1 points. We are asked to find the percentage of the data that is between 64.2 points and 88.6 points.

First of all, we will find z-scores of each data point as:

Let us find z-score corresponding to normal score 88.6.

To find the percentage of the data is between 64.2 points and 88.6 points, we need to find area under a normal distribution curve that lie within two standard deviation of mean.

The empirical rule of normal distribution states that approximately 95% of data points fall within two standard deviation of mean, therefore, option 'b' is the correct choice.

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

happy to

pls mark as brainliest.