Answer:
Step-by-step explanation:
Hello!
The variable of interest, X: height of women at a college, has an approximately normal distribution with mean μ= 65 inches and standard deviation σ= 1.5 inches.
You need to look for the value of height that marks the bottom 20% of the distribution, i.e. the height at the 20th percentile of the normal curve, symbolically:
P(X≤x₀)= 0.20
To know what value of height belongs to the 20% of the distribution, you have to work using the standard normal distribution and then reverse the standardization with the population mean and standard deviation to reach the value of X. So the first step is to look for the Z-value that accumulates 20% of the distribution:
P(Z≤z₀)=0.20
z₀= -0.842
z₀= (x₀-μ)/σ
z₀*σ= (x₀-μ)
x₀= (z₀*σ)+μ
x₀= (-0.842*1.5)+65
x₀= 63.737 inches
I hope it helps!
Answer:
Height at the 20th percentile = 63.74
Step-by-step explanation:
We are given that the heights of women at a college are approximately Normally distributed with a mean of 65 inches and a population standard deviation of 1.5 inches.
Since, X ~ N([tex]\mu=65,\sigma^{2}=1.5^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
As we are given that we have to find the height at the 20th percentile, which means the value of x below which the z has a area of 20% ;
P(X < x) = 0.20
P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-65}{1.5}[/tex] ) = 0.20
P(Z < [tex]\frac{x-65}{1.5}[/tex] ) = 0.20
Now, in z normal table, the critical value of x having area less than 20% is -0.8416, i.e;
[tex]\frac{x-65}{1.5}[/tex] = -0.8416
x = 65 - (0.8416 * 1.5)
x = 65 - 1.2624 = 63.74 inches
Therefore, height at the 20th percentile is 63.74 inches .
