Answered

Suppose baby kittens' weights are normally distributed with a mean of 10.2 and a standard deviation of 1.5. The [tex]$Z$[/tex]-score tells you how many units above the average (if the [tex]$Z$[/tex]-score is positive) or below the average (if the [tex]$Z$[/tex]-score is negative) any particular baby kitten's weight is.

Find the baby kitten weight that corresponds to the following Z-score using the formula [tex][tex]$Z=\frac{X-\mu}{\sigma}$[/tex][/tex], where [tex][tex]$\mu$[/tex][/tex] is the mean, [tex][tex]$\sigma$[/tex][/tex] is the standard deviation, and [tex][tex]$X$[/tex][/tex] is the baby kitten weight.

a. [tex][tex]$Z=1.49, X=$[/tex][/tex] [tex]$\square$[/tex]



Answer :

GinnyAnswer
To find the baby kitten's weight [tex]\( X \)[/tex] that corresponds to a [tex]\( Z \)[/tex]-score of 1.49, we can use the formula for the [tex]\( Z \)[/tex]-score:

[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

where:
- [tex]\( Z \)[/tex] is the [tex]\( Z \)[/tex]-score,
- [tex]\( \mu \)[/tex] is the mean of the dataset,
- [tex]\( \sigma \)[/tex] is the standard deviation, and
- [tex]\( X \)[/tex] is the value we want to find.

We are given the values:
- [tex]\( Z = 1.49 \)[/tex]
- [tex]\( \mu = 10.2 \)[/tex] (mean)
- [tex]\( \sigma = 1.5 \)[/tex] (standard deviation)

We need to rearrange the formula to solve for [tex]\( X \)[/tex]. Start by multiplying both sides by [tex]\( \sigma \)[/tex]:

[tex]\[ Z \cdot \sigma = X - \mu \][/tex]

Next, add [tex]\( \mu \)[/tex] to both sides to isolate [tex]\( X \)[/tex]:

[tex]\[ Z \cdot \sigma + \mu = X \][/tex]

Substitute the given values into the equation:

[tex]\[ 1.49 \cdot 1.5 + 10.2 = X \][/tex]

Perform the multiplication and addition:

[tex]\[ 2.235 + 10.2 = X \][/tex]

[tex]\[ X = 12.435 \][/tex]

Therefore, the baby kitten's weight that corresponds to a [tex]\( Z \)[/tex]-score of 1.49 is approximately [tex]\( 12.435 \)[/tex].

Other Questions