Complete Question
A population has a mean of 25, a median of 24, and a mode of 26. The standard deviation is 5. The value of the 16th percentile is _______. The range for the middle 3 standard deviation is _______.
Answer:
The value of the 16th percentile is [tex]x = 20[/tex]
The range for the middle 3 standard deviation is [tex]10 \to 40[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 25[/tex]
The median is [tex]m = 24[/tex]
The mode is [tex]n = 26[/tex]
The standard deviation is [tex]\sigma = 5[/tex]
Generally the 16th percentile is mathematically represented as
[tex]P(x)= P(\frac{X - \mu }{ \sigma } \le \frac{x- 25 }{5} ) = 0.16[/te
Generally [tex]\frac{X - \mu}{\sigma } = Z(The \ standardized \ value \ of \ X )[/tex]
So
[tex]P(x) = P(Z \le \frac{ x - 25}{5} ) = 0.16[/tex]
Now from the normal distribution table the z-score of 0.16 is
z = -1
[tex]P(x) = P(Z \le \frac{ x - 25}{5} ) = P(Z \le -1 )[/tex]
=> [tex]\frac{x- 25}{5} = -1[/tex]
=> [tex]x- 25 = -5[/tex]
=> [tex]x = 25-5[/tex]
=> [tex]x = 20[/tex]
The range for the middle 3 standard deviation is mathematically represented
[tex]\mu - 3\sigma \to \mu + 3\sigma[/tex]
[tex]25 - 3(5 ) \to 25 + 3(5 )[/tex]
[tex]10 \to 40[/tex]
