 Hello and welcome to the session. In this session, we will discuss a question which says that the reaction times for hand-eye coordination tests administered to 1,750 students are normally distributed with mean 0.45 seconds and a standard deviation of 0.05 seconds. About how many students had reaction time between 0.35 and 0.55 seconds? Now before starting the solution of this question, we should know some results. First is the class of normal distribution is symmetric about the mean, about 68% of data or area lies between x bar minus sigma and x bar plus sigma when about 95% of data or area lies between x bar minus 2 sigma and x bar plus 2 sigma and about 99% of data or area lies between x bar minus 3 sigma Now these results will work out as a key idea for solving out the given question. Now let us start with the solution of the given question. A test was conducted on 1,750 students. The average reaction time of students was 0.45 seconds and standard deviation was 0.05 seconds. So we are given mean that is x bar is equal to 0.45 and standard deviation that is sigma is equal to 0.05. Now it is also given to be a normal distribution. So we first draw a normal curve. To draw a normal curve we take horizontal axis. Now to draw a curve we find all the planes that is x bar minus sigma, x bar minus 2 sigma, x bar minus 3 sigma and x bar plus sigma, x bar plus 2 sigma, x bar plus 3 sigma. Now x bar minus sigma is equal to 0.45 minus 0.05 which is equal to 0.40. Then x bar plus sigma is equal to 0.45 plus 0.05 which is equal to 0.50. x bar minus 2 sigma is equal to 0.35, x bar plus 2 sigma is equal to 0.55. Then x bar minus 3 sigma is equal to 0.30 and x bar plus 3 sigma is equal to 0.60. So here we have written all these values on this horizontal line, the symmetry above the mean. So we have drawn vertical lines at these points keeping in mind the symmetry above the mean. And now we are drawing the top of these lines by freehand at reaction time between 0.35 and 0.55 seconds. For this we find the percentage of data lying in the internal 0.30.55. On the key idea we know the percentage distribution of area under the normal curve percentage of area between 0.35 and 0.55 seconds 13.5 percent plus 34 percent plus 34 percent plus 13.5 percent which is equal to 95 percent which means 95 percent of values lie between 0.35 seconds and 0.55 seconds which means 19 of students between 0.35 seconds and 0.55 seconds. Now total number of students is equal to 1 time means having reaction time between 0.35 and 0.550. Now this is equal to 95 upon 100 into 1750. Now 5 into 20 is 100 and 5 into 3 50 is 1750. Again 5 into 4 is 20 and 5 into 70 is 3 and this is equal to 3 upon 4 which is approximately the required answer during the session.