The burning times of scented candles, in minutes, are normally distributed with a mean of 249 and a standard deviation of 20. Find the number of minutes a scented candle burns if it burns for a shorter time than 60% of all scented candles.

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Answer:

The candle burns for 244 minutes.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 249, \sigma = 20[/tex]

Find the number of minutes a scented candle burns if it burns for a shorter time than 60% of all scented candles.

This is the 100-60 = 40th percentile, which is X when Z has a pvalue of 0.4. So X when Z = -0.253.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.253 = \frac{X - 249}{20}[/tex]

[tex]X - 249 = -0.253*20[/tex]

[tex]X = 244[/tex]

The candle burns for 244 minutes.