The lifespans of a species of fruit fly have a​ bell-shaped distribution, with a mean of 30 days and a standard deviation of 5 days.​(a) The life spans of three randomly selected fruit flies are 35 ​days, 25 ​days, and 42 days. Find the​ z-score that corresponds to each life span. Determine whether any of these life spans are unusual.​ (b) The life spans of three randomly selected fruit flies are 40 ​days, 25 ​days, and 20 days. Using the Empirical​ Rule, find the percentile that corresponds to each life span.​(a) The​ z-score corresponding a life span of 35 days is 1. The​ z-score corresponding a life span of 25 days is -1. The​ z-score corresponding a life span of 42 days is 2.4. ​(Type an integer or a decimal rounded to two decimal places as​ needed.)Select all of the life spans that are unusual. A. 35 daysB. 42 daysC. 25 daysD.None of the life spans are unusual. ​(b) Determine the percentiles using the Empirical Rule. The 40 day fruit fly corresponds to what percentile?The 25 day fruit fly corresponds to what percentile?The 20 day fruit fly corresponds to what percentile?

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We are told that the mean lifetime of a type of fruit fly is 30 days and that the distribution is bell-shaped. (This indicates that the distribution is approximately "normal" in the mathematical sense. The mean, median, and mode will be the same and the distribution will be symmetric about the mean.) The standard deviation of the lifetimes is 5 days.

(a) We are asked to compute the z-score for various lifetimes. The z-score gives the number of standard deviations above or below the mean for a given value. A negative z-score indicates a value below the mean, positive is above the mean, and a z-score of zero means the value is the mean.

The formula to compute the z-score is `z=(x-bar(x))/s` if we are dealing with a sample or `z=(x-mu)/sigma` if we are dealing with a population. In each case we take the difference between the data value and the mean divided by the standard deviation.

For example if x=35 then `z=(35-30)/5=5/5=1` . Thus 30 is 1 standard deviation above the mean. If x=25 then z=-1 and if x=42 then z=2.4. More than 2 standard deviations from the mean is atypical (as 95% of the population lies within 2 standard deviations from the mean) so 42 is unusual.

(b) The Empirical Normal Rule says that approximately 68% of the population lies within 1 standard deviation of the mean. (Since the distribution is assumed to be symmetric this means that 34% lie within 1 standard deviation above the mean and 34% below.)

95% of the population lie within 2 standard deviations of the mean. (Thus 13.5% lie between 1 and 2 standard deviations above the mean, and 13.5% between 1 and 2 below.)

99.7% lie within 3 standard deviations of the mean. (Then 2.35% lie between 2 and 3 standard deviations above the mean, 2.35% between 2 and 3 below.)

This leaves .3% outside of 3 standard deviations from the mean; .15% above and .15% below.

So for x=40 we find the z-score to be 2. This value lies 2 standard deviations above the mean. This means that the 50% of the population below the mean are below that score, as well as the 34% that are within 1 standard deviation of the mean, and the 13.5% that lie between 1 and 2 standard deviations above the mean. Thus the percentile for x=40 is the 97.5th percentile. (We could also see that only .0235+.0015=.025 or 2.5% of the population scores higher.

In a similar vein of x=25, the z-score is -1, and the percentile is the 16th percentile. For x=20, the z-score is -2, and the percentile is the 2.5th percentile.

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