Compute Mean of a Clipped Normal Distribution I am trying to solve this problem and have solved all parts except for the last part.
I have tried to research the problem and saw some theory on truncated normal distributions and found this formula on Wikipedia
Which seems to be of expected value and I am assuming this would be equal to median in a normal distribution scenario, but it does not give the required answer.  The answer of part b is (351)
I have two questions:

*

*please could someone help me solve part (b)?

*Is median=mean assumption correct in a truncated normal distribution scenario?

 A: 
Is median=mean assumption correct in a truncated normal distribution
scenario?

No, because the truncated distribution may be asymmetric.

please could someone help me solve part (b)?

It's easy :)
The proportion of candidates that take no longer than 450 seconds is $P(X\le 450)$ where $X\sim\mathcal{N}(360,75)$, $88.49\%$. In R:
> (b <- pnorm(450,360,75))
[1] 0.8849303

The proportion of candidates that take no longer than 210 seconds is $P(X\le 210)=2.275\%$:
> (a <- pnorm(210,360,75))
[1] 0.02275013

The proportion of candidates that are interviewed is $86.22\%$:
> b-a
[1] 0.8621802

They are between the $a=0.02275$ and the $b=0.8849$ percentiles, so you need the value $X=x$ corrensponding to the $(b-a)/2+a$ percentile:
> (c <- qnorm((b-a)/2+a,360,75))
[1] 351.3026

Indeed:
> pnorm(450,360,75)-pnorm(c,360,75)
[1] 0.4310901
> pnorm(c,360,75)-pnorm(210,360,75)
[1] 0.4310901

A: Graphical comment: You have a fine answer from @Sergio (+1). Here is a simulation in R of an imaginary
one million original applicants, which gives approximate answers and makes
it easy to plot relevant histograms.
hist(y, prob=T, br=30, col="skyblue2", main="Interviewed")
set.seed(2020)
x = rnorm(10^6, 360, 75)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  6.994 309.480 359.958 360.041 410.543 722.051 
sd(x)
[1] 74.9409

hist(x, prob=T, br=50, col="skyblue2", main="Sample from NORM(360, 75)")
 curve(dnorm(x, 360, 75), add=T, lwd=2, col="purple")
 abline(v=c(210,450), col="darkgreen")

Interviewed applicants have times between the vertical green lines. The density function of $\mathsf{Norm}(\mu=360, \sigma=75)$ is shown along with the histogram.

Now we isolate the 862,126 applicants who will be interviewed. Their median
time is about 351 (vertical red line on histogram below). Half of the interviewees had times on either side of this line. This is not a symmetrical distribution. Its mean is at
about 348. A density curve of the truncated distribution is shown along with the histogram.
y=x[x > 210 & x < 450]
summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  210.0   306.6   351.3   347.8   392.8   450.0 
sd(y)
[1] 56.98087
length(y)
[1] 862126

hist(y, prob=T, br=30, col="skyblue2", main="Interviewed")
 DF = diff(pnorm(c(210,450), 360, 75))
 curve(dnorm(x, 360, 75)/DF, add=T, lwd=2, col="purple")
 abline(v=quantile(y,.5), col="red")


