Calculating the random variable for 99% confidence interval correct, understand, like, question, standard

kseniaru

Hi, i have a question and i like to understand how to calculate.

An analyst determined that the sample mean and variance for a normal distribution are 42 and 9 respectively. The 99% confidence interval for this random variable is closest to:

a) 15.0 to 69.0
b) 18.8 to 65.2
c) 34.3 to 49.7
d) 39.0 to 45.0

The correct ans is c.

The 99% confidence interval for a normally distributed random variable is equal to
the sample mean +/- (2.58 * sample standard deviation)

= 42 +/- (2.58*90.5)
= 42 +/- (2.58 * 3)
= 42 +/- 7.74
which is 34.3 to 49.7

Actually how does is derive the value 7.74? How is it calculate? thanks

Siddimaula

Oh thanks for the explanation...