How can i calculate probabilities regarding candle factories? (details inside)
hello, i'm having trouble with this question. would appreciate your help with it:
in a package of candles there are 45 candles, with random sizes(lengths) and no dependency between the lengths.
in a certain factory, candles are produced so that the distribution of the length (in cm) (of each) is normal with the parameters 13 and 0.1^2
1)what is the probability that in a random package there will be at least 30 candles whose length is within the range of 12.82 and 13.06 cm?
2)what is the length of the candle that 92% of the candles are shorter than him?
hello, i'm having trouble with this question. would appreciate your help with it:
in a package of candles there are 45 candles, with random sizes(lengths) and no dependency between the lengths.
in a certain factory, candles are produced so that the distribution of the length (in cm) (of each) is normal with the parameters 13 and
1)what is the probability that in a random package there will be at least 30 candles whose length is within the range of 12.82 and 13.06 cm?
2)what is the length of the candle that 92% of the candles are shorter than him?
1 Answer
1) 0.6903
2) 13.141 cm
Explanation:
Let
1)
We first find the probability of selecting a single candle that's between 12.82 and 13.06 cm.
"P"(12.82 < X < 13.06)
= "P"((12.82-13)/0.1 < Z < (13.06-13)/0.1)
= "P"(–1.8 < Z < 0.6)
= "P"(Z < 0.6) - "P"(Z < –1.8)
= 0.7257 - 0.0359
=0.6898
This is the probability of "success" for a single candle.
Let
We now seek
"P"(Y "=" 30) + "P"(Y "=" 31) + ... + "P"(Y "=" 45)
This would take a long time. However, since
Y " "stackrel "approx." ~ "N"(np, npq) " "="N"(31.041, 9.6289)
Using a continuity correction, we get
stackrel"Binomial" overbrace("P"(Y >= 30)) ~~ stackrel"Normal" overbrace("P"(Y >= 29.5))
color(white)("P"(Y >= 30)) = 1- "P"(Y < 29.5)
color(white)("P"(Y >= 30)) = 1- "P"(Z < (29.5-31.041)/sqrt9.6289)
color(white)("P"(Y >= 30)) = 1- "P"(Z < –0.4966)
color(white)("P"(Y >= 30)) = 1- 0.3097" " (from software)
color(white)("P"(Y >= 30)) = 0.6903
The actual Binomial probability is
2)
We seek
"P"(Z < (x - mu)/sigma) = 0.92
Through table lookup, we get
(x - mu)/sigma = z ~~ 1.41
:. (x - 13)/0.1 ~~ 1.41
" "x - 13 ~~0.141
" "x ~~13.141
So 92% of the candles will be shorter than 13.141 cm.