A certain animal's body temperature has a mean of 94.72° F and a standard deviation of 0.57°F. Convert the given temperatures to z scores.
a. 93.52 °F b. 95.22 °F c. 94.72 °F
Answer:
a. z = - 2.1053
b. z = 0.87719
c. z = 0
Step-by-step explanation:
Given that :
The population mean μ = 94.72
The standard deviation σ = 0.57
the formula for calculating the standard normal z score, which can be represented as:
[tex]z= \dfrac{\overline x - \mu}{\sigma}[/tex]
For a.
The sample mean [tex]\bar x[/tex] = 93.52
The z score can be computed as follows:
[tex]z= \dfrac{\overline x - \mu}{\sigma}[/tex]
[tex]z= \dfrac{93.52 - 94.72}{0.57}[/tex]
[tex]z= \dfrac{-1.2}{0.57}[/tex]
z = - 2.1053
For b.
The sample mean [tex]\bar x[/tex] = 95.22
[tex]z= \dfrac{\overline x - \mu}{\sigma}[/tex]
[tex]z= \dfrac{95.22 - 94.72}{0.57}[/tex]
[tex]z= \dfrac{0.5}{0.57}[/tex]
z = 0.87719
For c.
The sample mean [tex]\bar x[/tex] = 94.72
[tex]z= \dfrac{\overline x - \mu}{\sigma}[/tex]
[tex]z= \dfrac{94.72 - 94.72}{0.57}[/tex]
[tex]z= \dfrac{0}{0.57}[/tex]
z = 0
