Question

A normal distribution of data has a mean of 90 and a standard deviation of 18. what is the approximate z-score for the value 64? –3.6 –1.4 1.4 3.6

Answer 1

-1.4

Explanation:

Answer 2

The approximate z-score for the value 64 is -1.4.

Explanation:

Since, the z-score or standard score for the value x is,

`z=(x-\mu)/(\sigma)`

Where,

`\mu` is the mean score,

`\sigma` is the standard deviation,

Here, x = 64,

`\mu=90`

`\sigma=18`

Hence, the z-score for the value 64 is,

`z=(64-90)/(18)`

`=(-26)/(18)`

`=-1.44444444444\approx -1.4`

Second option is correct.