If $X$X is a normally distributed set of scores with mean = $\mu$μ and variance = $\sigma^2$σ2 then we can write this as $X\sim N(\mu,\sigma^2)$X~N(μ,σ2). For example; $X\sim N(6,100)$X~N(6,100) means $X$X is a set of scores, normally distributed, with mean = $6$6 and variance = $100$100 (so standard deviation is $10$10) .
The standard normal distribution has mean of $0$0 and variance of $1$1. It is denoted with the variable $Z$Z. We can write $Z\sim N(0,1)$Z~N(0,1).
$P(Z<2.2)$P(Z<2.2) means determine the probability that a score in a standard normal distribution is less than $2.2$2.2. It can also be interpreted as find the area under the standard normal curve to the left of $2.2$2.2.We saw in our previous chapter how we can use the graph of the normal distribution and the $68-95-99.7%$68−95−99.7% rule to calculate probabilities for both the standard normal distribution where $\mu=0$μ=0 and $\sigma=1$σ=1, and also the general normal distribution, where $\mu$μ and $\sigma$σ take on any value.
When we wish to calculate probabilities for values that are not exactly $1$1, $2$2 or $3$3 standard deviations from the mean, we turn to technology to perform our calculations.
Let's take a look at how to do these using our calculator.
Casio ClassPad
How to use the CASIO Classpad to complete the following tasks involving calculating probabilities for the normal distribution.
Use your calculator to calculate the following probabilities. Give answers correct to $4$4 decimal places.
$P\left(Z<2.2\right)$P(Z<2.2)
$P\left(-1.2
$P\left(X>90\right)$P(X>90) where $X~N\left(100,7^2\right)$X~N(100,72)
$P\left(12
TI Nspire
How to use the TI Nspire to complete the following tasks involving calculating probabilities for the normal distribution.
Use your calculator to calculate the following probabilities. Give answers correct to $4$4 decimal places.
$P\left(Z<2.2\right)$P(Z<2.2)
$P\left(-1.2
$P\left(X>90\right)$P(X>90) where $X~N\left(100,7^2\right)$X~N(100,72)
$P\left(12
Using your calculator, find the probability that a $z$z-score is at most $1.60$1.60 given that it is greater than $-0.69$−0.69 in the standard normal distribution.
Give your answer correct to $4$4 decimal places.
If $X\sim N\left(20,5^2\right)$X~N(20,52), use your calculator to determine $k$k in the following parts.
Round your answers to two decimal places.
$P\left(X
$P\left(X>k\right)=0.45$P(X>k)=0.45
$P\left(k
$P\left(21
What does it mean when you hear that someone is in the $98$98th percentile?
Have you ever completed a mathematics competition and when you receive your result you find that you were above the $0.85$0.85 quantile?
You know this is a good thing, but do you know what it means?
As the name suggests, percentiles split an ordered set of data into one hundred parts, where each percentile indicates the proportion of the population below that value.
Let’s think about the height of female adults. If the $45$45th percentile height is $155$155 cm, this means that $45%$45% of female adults are shorter than $155$155 cm.
We often see our test results expressed in this way: “You are placed in the $78$78th percentile”. But what does this mean?
If you are at the $78$78th percentile, then you performed better than $78%$78% of those who did the test.
But if you are in the $78$78th percentile, then you performed better than at least $78%$78% of those who did the test.
In the second case, we can say "at least" $78%$78% because you may have scored better than other students also in the $78$78th percentile. But in both cases your score is still below the percentiles above the $78$78th (i.e. the $79$79th, $80$80th, etc up to the $99$99th percentile). This is illustrated in the images below:
Note that the percentile is not the same as your actual score in the test. Rather it is a reflection of your rank in the ordered set of scores from all the participants.
A quantile is exactly like a percentile, but expressed as a decimal instead.
So for the situation above, where you're in the $45$45th percentile for your height, you would say you're in the $0.45$0.45 quantile.
If you're in the $94$94th percentile for your mathematics competition result, then you're in the $0.94$0.94 quantile.
The results of an IQ test are known to be normally distributed with a mean of $100$100 and a standard deviation of $10$10.
(a) What is the $84$84th percentile?
Think: We need to consider how many standard deviations above or below the mean indicates that an area of $84%$84% has been shaded on our normal distribution curve. Thinking about $84%$84%, we know this is made up of $50%$50% (half the curve) and $34%$34%. From our experience with the $68-95-99.7%$68−95−99.7% rule, we know that $34%$34% means we have one standard deviation above the mean. We can sketch this on a graph.
Do:
From our graph, we can see that the $84$84th percentile is $110$110 on this IQ test.
(b) What is the lowest mark achieved by the $0.975$0.975 quantile?
Think: The lowest mark achieved by the $0.975$0.975 quantile is the same as the mark that is at the $0.975$0.975 quantile. Using similar reasoning as we did in part (a), we know that $0.975$0.975 is made up of $0.5$0.5 (half the curve) and $0.475$0.475 which is half of $0.95$0.95. Thus we're looking at the score that is two standard deviations above the mean. Again, a graph is useful here.
Do:
From our graph, we can see that the lowest mark achieved by the $0.975$0.975 quantile is $120$120 on the IQ test.
Let's look at an example using the inverse normal function of our calculator to find scores given a probability. Select your calculator brand below to work through an example.
Casio ClassPad
How to use the CASIO Classpad to complete the following tasks involving the inverse normal distribution.
The heights of $16$16-year-old females are known to be normally distributed with a mean of $165$165 cm and a standard deviation of $2$2 cm.
Give your answers correct to $2$2 decimal places.
Calculate the height that $98%$98% of $16$16-year-old females fall below.
What is the height of the $0.4$0.4 quantile?
What is the shortest height of the tallest $15$15% of females?
TI Nspire
How to use the TI Nspire to complete the following tasks involving the inverse normal distribution.
The heights of $16$16-year-old females are known to be normally distributed with a mean of $165$165 cm and a standard deviation of $2$2 cm.
Give your answers correct to $2$2 decimal places.
Calculate the height that $98%$98% of $16$16-year-old females fall below.
What is the height of the $0.4$0.4 quantile?
What is the shortest height of the tallest $15$15% of females?
For the standard normal variable $X$X$~$~$N\left(0,1\right)$N(0,1), use a graphics calculator to determine the following values.
Round your answers to three decimal places.
The $0.7$0.7 quantile
The $65$65th percentile
The lowest score in the top $20$20 percent
If $X\sim N\left(30,4^2\right)$X~N(30,42), determine:
the $0.5$0.5 quantile.
the $0.83$0.83 quantile.
Round your answer to two decimal places.
the $35$35th percentile.
Round your answer to two decimal places.
Since we can rely on technology to calculate probabilities for any general normal distribution, we rely less on the use of the standard normal distribution than we did in the past. However, when we don't know the mean and/or the standard deviation of a normal variable $X$X, we can use the standard normal distribution $Z$Z to help us calculate these.
If $X\sim N\left(\mu,100\right)$X~N(μ,100), use your calculator to find $\mu$μ if $P\left(\mu\le X\le20\right)=0.3013$P(μ≤X≤20)=0.3013
Round your answer to two decimal places.
If $X\sim N\left(\mu,\sigma^2\right)$X~N(μ,σ2), use your calculator to find $\mu$μ and $\sigma$σ if $P\left(X<70\right)=0.1817$P(X<70)=0.1817 and $P\left(X<80\right)=0.9655$P(X<80)=0.9655
Round your answers to two decimal places.
$\mu$μ | $=$= | $\editable{}$ |
$\sigma$σ | $=$= | $\editable{}$ |