A normal distribution of raw scores has a mean of 50 and a s

A normal distribution of raw scores has a mean of 50 and a standard deviation of 7. Approximately 95% of the distribution can be found between what two raw scores? A. 36 and 50 B. 50 and 64 C. 36 and 64 D. 43 and 57

Solution

The 95 % confidence interval of a normal distribution is [ -1.96 , + 1.96] where is the mean and , the standard deviation. Here, = 50 and = 7 so that 1.96 = 7*1.96 = 13.72. Thus, the 95% confidence interval for the given distribution is [ 50- 13.72, 50 + 13.72] i.e. [ 36.28, 63.72] or, [ 36, 64] on rounding off to the nearest whole numbers.. The answer C is correct.