Question

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?

Answer 1

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.