Question

Suppose "x" is a normal random variate with mean 5 and variance 16
the probability of X<1 in terms of the CDF of the standard normal random variate is

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Answer 1

Given :   "x" is a normal random variate with mean 5 and variance 16

To Find : probability of x<1

Solution:

Mean = 5

Variance = 16

Variance = SD²

=> SD = 4

SD is standard deviation

Z score = ( Value - Mean ) /SD

x = 1

=> Z score =  ( 1 - 5) / 4 = - 1

from Z score  table :

0.1587   are less than x = 1

Hence probability of x<1 is 0.1587

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Answer 2

Given:-

  "x" is a normal random variate with mean 5 and variance 16.

_____________________

To Find:-

• probability of x<1

______________________

step-by-step solution:-

→ Mean = 5

→ Variance = 16

→ Variance = SD²

=> SD = 4

SD is standard deviation.

→ Z score = ( Value - Mean ) /SD

x = 1

→ Z score =  ( 1 - 5) / 4 = - 1

from Z score  table :

0.1587   are less than x = 1

Hence, probability of x<1 is 0.1587