snowlovelydayM

2021-02-08

Find the following percentiles for the standard normal distribution. Interpolate where appropriate.
a.91st
Percentage value is 91.

Tasneem Almond

Calculation:
Denote the (100p) th percentile of the standard normal distribution as n(p).
It is known that if the (100p) th of the distribution of a random variable is n(p), then, $P\left(X\le n\left(p\right)\right)=p$, that is, $\varphi \left(n\left(p\right)\right)=p$.
Here,
$100p=91$
$p=091$.
Thus, the corresponding percentile is $\varphi \left(n\left(0.91\right)\right)=0.91$.
In Table A-3 “Standard Normal Curve Areas” find the value of the variable which corresponds to the probability 0.9100.
Procedure:
Locate the probability value 0.9100 in the body of the table.
The exact value 0.9100 is not available. However, the probability 0.9099 is available, which is pretty close to 0.9100. That is, $0.9100\sim 0.9099$.
Move horizontally from the probability value till the first column is reached and note the value of z as 1.3.
Move vertically from the probability value till the first row is reached and note the value of z as .04.
Thus,
$\varphi \left(1.34\right)=0.9099$
0.9100.
Now, $\varphi \left(n\left(0.91\right)\right)=0.91$. Hence, the required percentile for the standard normal distribution is approximately 1.34.

user_27qwe

Step 1:
We can make use of z-scores, which measure the number of standard deviations a given value is from the mean of the distribution. The z-score formula is defined as:
$z=\frac{x-\mu }{\sigma }$
where $x$ is the value we want to find, $\mu$ is the mean of the distribution (which is 0 for the standard normal distribution), and $\sigma$ is the standard deviation of the distribution (which is 1 for the standard normal distribution).
Step 2:
Since the problem asks for the 91st percentile, we need to find the value of $x$ for which the area to the left of $x$ on the distribution curve is 0.91.
To find this value, we can use a standard normal distribution table or a statistical software.
1. Convert the given percentage value to a decimal by dividing it by 100:

2. Look up the corresponding z-score for the desired percentage value of 0.91 in the standard normal distribution table or using a statistical software. In this case, the z-score is approximately 1.34.
3. Substitute the known values into the z-score formula and solve for $x$:
$1.34=\frac{x-0}{1}$
Simplifying the equation gives:
$x=1.34×1=1.34$
Therefore, the value corresponding to the 91st percentile for the standard normal distribution is approximately 1.34.

karton

To find the 91st percentile for the standard normal distribution, we can use the cumulative distribution function (CDF) of the standard normal distribution, denoted by $\Phi \left(x\right)$.
The 91st percentile corresponds to the value $x$ for which $\Phi \left(x\right)=0.91$. We can solve for $x$ by finding the inverse of the CDF, denoted as ${\Phi }^{-1}\left(p\right)$, where $p$ is the desired percentile.
Therefore, to find the 91st percentile, we need to calculate ${\Phi }^{-1}\left(0.91\right)$.

star233

Explanation:
The standard normal distribution is denoted by $Z$ and has a mean ($\mu$) of 0 and a standard deviation ($\sigma$) of 1. The percentile value we want to find is 91%.
First, we find the corresponding $z$-score using the cumulative distribution function (CDF) of the standard normal distribution. The CDF gives us the probability of obtaining a value less than or equal to a given $z$-score.
Let's denote the $z$-score for the 91st percentile as ${z}_{0.91}$. We can find it using the inverse CDF, also known as the quantile function or percent-point function.
To find ${z}_{0.91}$, we have:
${z}_{0.91}={\Phi }^{-1}\left(0.91\right)$
Using a standard normal distribution table or a calculator, we can determine that ${\Phi }^{-1}\left(0.91\right)\approx 1.34$.
Therefore, the $z$-score corresponding to the 91st percentile is approximately 1.34.

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