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?

1 Answer
Geoff K.
May 27, 2018

1) 0.6903

2) 13.141 cm

Explanation:

Let #X# be the length of a single candle. Then #X" ~ N"(13, 0.1^2).#

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 #Y# be the number of candles out of 45 that are between 12.82 and 13.06 cm. Then #Y" ~ Bin(45, 0.6898)."#

We now seek #"P"(Y >= 30)#. Usually that would mean calculating

#"P"(Y "=" 30) + "P"(Y "=" 31) + ... + "P"(Y "=" 45)#

This would take a long time. However, since #n=45# is fairly large and #p=0.6898# is fairly close to #0.5,# we can use the normal approximation for #Y.#

#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 #"P"(Y <= 30) ~~ 0.6955# so our Normal approximation is pretty good.

2)

We seek #x# such that #"P"(X < x) = 0.92#. This is the same as

#"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.

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