Answer :
Answer:
The 9th percentile is 40.52.
Step-by-step explanation:
We are given that the random variable X is normally distributed with a mean of 50 and a standard deviation of 7.
Let X = the random variable
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 50
[tex]\sigma[/tex] = standard deviation = 7
So, X ~ Normal([tex]\mu=50, \sigma^{2} = 7^{2}[/tex])
Now, the 9th percentile is calculated as;
P(X < x) = 0.09 {where x is the required value}
P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-50}{7}[/tex] ) = 0.09
P(Z < [tex]\frac{x-50}{7}[/tex] ) = 0.09
Now, in the z table the critical value of x that represents the below 9% of the area is given as -1.3543, i.e;
[tex]\frac{x-50}{7}=-1.3543[/tex]
[tex]x-50=-1.3543 \times 7[/tex]
[tex]x=50 -9.48[/tex]
x = 40.52
Hence, the 9th percentile is 40.52.