Using the information from the histogram and assuming that this measure of confidence in institutions is normally distributed. Determine the probability that a randomly drawn respondent has Y confidence in institutions where Y:
Histogram: https://i.stack.imgur.com/SKrEJ.jpg
⦁ is greater than or equal to 8
⦁ is less than or equal to 5
⦁ lies between 7 and 9
⦁ Estimate the score equal to p90
I thought to use z-scores here so then:
Mean = 6
sd = 2
for the first part:
z-score = (8-6)/2 = 1
p(z>1) 1 - p(z<1) = 1 - .8413 = .1587 or 15.87percent
for the second part:
z-score = (5-6)/2 = -1/2 = -0.5
p(z<-0.5) = .3085 or 30.85percent
for the third part:
z-score = 7-6/2 = 0.5 = .6915 z-score = 9-6/2 = 1.5 = .9332 p(7 < z < 9) = .9332 - .6915 = .2417 or 24.17percent
Can someone please tell me if what I've done is incorrect or correct.
Also confused on how to estimate the score equal to P90 percentile.
Please help Thank you
Hints
ad a) First of all this is a discrete distribution. That means $P(X\geq 8)=1-P(X\leq 7)$. And therefore you need additionally the continuity correction factor $0.5$.
$$P(X\geq 8)\approx 1-\Phi\left( \frac{7+0.5-6}{2} \right)=1-\Phi\left( 0.75\right) $$
Using this online calculator we get $P(X\geq 8)\approx 1-0.773=0.227=22.7\%$. Note that this is an approximation.
Similar for b)-add the continuity correction factor.
ad c) If we interpete this as $P(7\leq X\leq 9)$ then you have to calculate
$P(7\leq X\leq 9)=P(X\leq 9)-P(X\leq 6)$