Question

68 % of students at a school weight between 54 kg and 86 kg. Assuming this data is normally distributed, what are the mean and standard deviation?

Answer 1

Let X be a random variable denoting the weight of the students which follows normal distribution with mean and standard deviation .

Let the mean and standard deviation are and .

Now it is given that 68% of students at a school weight between 54 kg and 86 kg.

Here, P(54<X<86)=0.68.

P[(54-)/<(X-)/<(86-)/]=0.68

P[(54-)/<Z<(86-)/]=0.68 [where Z=(X-)/ is a standard normal variable with mean 0 and standard deviation 1].

- =0.68.....(1)

Now since Z follows a normal distribution, it is symmetric. Hence, (54-)/ and (86-)/ will be in the same distance but in opposite sides from the axis of symmetry. Thus, (86-)/=- (54-)/=70.

Now equation (1) can be written as-

2*-1=0.68=0.84

Now putting =70 from the standard normal distribution table we get, (86-70)/=0.99/16=1/0.99=1616.

Therefore the mean and the standard deviation of X are 70 and 16.16 respectively.