Question

The life span of a species of fruit fly have a bell-shaped distribution, with a mean of 33 days and a standard deviation of 4 days.

What percentage corresponds to the life span of fruit flies that are between 29 days and 37 days?

Answer 1

To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores. The percentage is approximately 68%.

Step-by-step explanation:

To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores.

To calculate the z-scores, we use the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

For 29 days: z = (29 - 33) / 4 = -1

For 37 days: z = (37 - 33) / 4 = 1

Looking at the standard normal distribution table or using a calculator, we find that the area between -1 and 1 is approximately 68%. Therefore, the percentage that corresponds to the life span of fruit flies between 29 days and 37 days is 68%.

Answer 2

68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days

Step-by-step explanation:

Given that the life span of a species of fruit fly have a bell-shaped distribution, with a mean of 33 days and a standard deviation of 4 days.

Let X be the life span

X is N(33,4)

we can convert X into Z standard normal variate by

`z=(x-33)/(4)`

To find the percentage corresponds to the life span of fruit flies that are between 29 days and 37 days

For this let us find probability for x lying between 29 and 37 days

`29\leq x\leq 37\\=-1\leq z\leq 1`

Probability = P(|z|<1) = 0.6826

Convert to percent

68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days