The continuous random variable, $X$, is normally distributed with$P(X<55) = 0.7$
a) How many standard deviations from the mean is a score of $55$?
b) If the standard deviation of $X$ is $4$, find the mean of the distribution.
For part a, the textbook solves $P(Z≤a) = 0.7$, giving the answer $0.5244$. I don't understand how this works. From my understanding, the standard normal distribution has mean of $0$ and S.D of $1$. But in this case, we are only given that the probability of obtaining a score of less than $55$ is $0.7$ - how do you find the number of standard deviations if I don't know what the mean is in this particular instance. Don't general normal distributions, not standard, vary in their measure of mean and S.D?
This is a high school question covered under the "normal distribution" topic.
If anyone could provide an explanation, it would be much appreciated!
$\endgroup$2 Answers
$\begingroup$We know $P(X<x)=\phi(x)$
So,
$P\bigg(\dfrac{X-\mu}{\sigma}<\dfrac{55-\mu}{\sigma}\bigg)=0.7 \implies \phi\bigg(\dfrac{55-\mu}{\sigma}\bigg)=0.7$
$P(X<a)=\phi(a)=0.7$
So for what value of $a$ area is $0.7$?
As you have written $a$ is $0.5244$
Also : $\dfrac{55-\mu}{\sigma}=0.5244$
"how do you find the number of standard deviations if I don't know what the mean is in this"
Explanation: The standard normal distribution is centered at mean=zero and the degree to which a given measurement deviates from the mean is given by the standard deviation. For the standard normal distribution, $68$% of the observations lie within $Z=1$ standard deviation of the mean; $95$% lie within two standard deviations of the mean, and $99.9$% lie within $Z=3$ standard deviations of the mean.
So your Z-Score values tell you how much away you are from the mean in our case we are $0.5244$ standard deviation away from the mean.
You can play around with normal curve from here: Pic source : Here
$\endgroup$ $\begingroup$Let us say that $X$ has mean $\mu$ and standard deviation $\sigma$.
Then $0.7=P\left(X<55\right)=P\left(\frac{X-\mu}{\sigma}\leq\frac{55-\mu}{\sigma}\right)$
Here $U:=\frac{X-\mu}{\sigma}$ has standard normal distribution andwith an appropriate table on standard normal distribution we can find the value $u$ that satisfies:$$0.7=P\left(U<u\right)$$
(as it seems $u\approx0.5244$, but I will leave that aside)
Then we can conclude that $\frac{55-\mu}{\sigma}=u$ or equivalently$55=\mu+\sigma u$.
So $u$ is here the answer on question a).
If $\sigma=4$ then we find $\mu=55-4u$ as answer on question b)
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