I\'m having trouble with the second portion of the question. If anyone could pos
ID: 3262429 • Letter: I
Question
I'm having trouble with the second portion of the question. If anyone could possibly tell me how to figure this out with statcrunch or manually, I would appreciate it
Explanation / Answer
Here X is the reading on thermometer. So X~N(0,1002)
Let p be the 84th percentile, i.e P84 = p . (Now percentile means dividing the region into 100 parts)
For the graph, the answer will be D as it show approximately 84% of the region i.e the probabilty where P(X<p)=0.84
From the Biometrica table we can get that 0.8413 is the approx Z-value of 1.00. (Z represents standard normal variable, Z~N(0,1) ) .So, P[z<1.00]= 0.8413
So,
P(X<p) =0.84
=> P[(X/100)<(p/100)]= P[z<1.00] (Standardizing X )
=> P[z<(p/100)]= P[z<1.00] ( In general Z=(X-mean)/S.D )
hence, p/100= 1.00
=> p=100
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