Compute Mean of a Clipped Normal Distribution

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")

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 $P(Xle 450)$ where $Xsimmathcal{N}(360,75)$, $88.49%$. In R:

> (b <- pnorm(450,360,75))
[1] 0.8849303

The proportion of candidates that take no longer than 210 seconds is $P(Xle 210)=2.275%$:

> (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