Question

Researchers recorded the speed of ants on trails in their natural environments. The ants studied, Leptogenys processionalis, all have the same body size in their adult phase, which made it easy to measure speeds in units of body lengths per second (bl/s). The researchers found that, when traffic is light and not congested, ant speeds vary roughly Normally, with mean 6.20 bl/s and standard deviation 1.58 bl/s. (a) What is the probability that an ant's speed in light traffic is faster than 5 bl/s? You may find Table B useful. (Enter your answer rounded to four decimal places.) probability: (b) What percent of ant speeds in light traffic are slower than 6 bl/s? (Enter your answer rounded to two decimal places.) percent c) What are the 10th and 90th percentiles of ant speeds in light traffic? (Enter your answers rounded to four decimal places.) 10th percentile bl/s 90th percentile: bl/s

Answer 1

Answer:

(a). Probability is 0.7764; (b) 44.83%; (c) The 10th percentile is x = 4.1776 bl/s and 90th percentile is x = 8.2224 bl/s.

Step-by-step explanation:

A short introduction

Standard normal table

To answer these questions, we need to use the standard normal table--the probabilities related to all normally distributed data can be obtained using this table, regardless of the population's mean and the population standard deviation.

Raw and z-scores

For doing this, we have to 'transform' raw data into z-scores, since the standard normal table has values from the cumulative distribution function of a standard normal distribution.

The formula for z-scores is as follows:

$\\ z = \frac{x - \mu}{\sigma}$ (1)

Where

$\\ x\;is\;the\;raw\;score$.

$\\ \mu\;is\;the\;population\;mean$.

$\\ \sigma\;is\;the\;population\;standard\;deviation$.

As we can see, the z-scores 'tell' us how far are the raw scores from the mean. Likewise, we have to remember that the normal distribution is symmetrical. It permits us, among other things, to find that certain scores (raw or z) have symmetrical positions from the mean and use this information to find probabilities easier.

Percentiles

Percentiles are values or scores that divide the probability distribution into two parts. For instance, a 10th percentile is a value in the distribution where 10% of the cases are below it, and therefore 90% of them are above it. Conversely, a 90th percentile is the value in the distribution where 90% of the cases are below it and 10% of them are above it.

Parameters of the distribution

The mean of the distribution in this case is $\\ \mu = 6.20$.

The standard deviation of the distribution is $\\ \sigma = 1.58$.

Having all this information, we can start solving the questions.

Probability that an ant's speed in light traffic is faster than 5 bl/s

For a raw score of 5 bl/s, the z-score is

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{5 - 6.20}{1.58}$

$\\ z = -0.759493 \approx -0.76$

As we can see, this value is below the mean because of the negative sign.

In general, the cumulative standard normal table does not work with negative values. To overcome this, we take advantage of the symmetry of the normal distribution. For a z = -0.76, and because of the symmetry of the normal distribution, the score z = 0.76 is above the mean (the positive sign indicates this) and is symmetrically positioned. The difference is that the cumulative probability is greater but the complement (P(z>0.76)) is the corresponding cumulative probability for z = -0.76. Mathematically:

$\\ P(z$

So

Consulting the cumulative standard normal table, for P(z<0.76) = 0.77637.

Then

$\\ P(z$

$\\ P(z0.76)$

$\\ P(z0.76)$

Rounding to four decimal places, the P(z<-0.76) = 0.2236 or 22.36%.

But this is for ants that are slower than 5 bl/s. For ants faster than 5 bl/s, we have to find the complement of the probability (symmetry again):

$\\ P(z>-0.76) = 1 - P(z$

Then, the probability that an ant's speed in light traffic is faster than 5 bl/s is 0.7764.

Percent of ant speeds in light traffic is slower than 6 bl/s

We use the same procedure as before for P(x<6 bl/s).

The z-score for a raw value x = 6 is

$\\ z = \frac{x - \mu}{\sigma}$

$\\ z = \frac{6 - 6.20}{1.58}$

$\\ z = -0.12658 \approx -0.13$

The value is below and near the population mean.

For a z = 0.13, the cumulative probability is 0.55172.

Then

$\\ P(z$

$\\ P(z0.13)$

$\\ P(z0.13)$

Or probability is 0.44828. Rounding to two decimal places P(z<-0.13) = 44.83%.

So, the percent of ant speeds in light traffic that is slower than 6 bl/s is 44.83% (approximately).

The 10th and 90th percentiles  

These percentiles are symmetrically distributed in the normal distribution. Ten percent of the data of the distribution is below the 10th percentile and 90% of the data is below the 90th percentile.

What are these values? We have to use the formula for z-scores again, and solve the equation for x.

For a probability of 90%, z is 1.28 (approx.)

$\\ 1.28 = \frac{x - 6.20}{1.58}$

$\\ x = (1.28 * 1.58) + 6.20$

$\\ x = 8.2224\frac{bl}{s}$

By symmetry, for a probability of 10%, z is -1.28 (approx.)

$\\ -1.28 = \frac{x - 6.20}{1.58}$

$\\ x = (-1.28 * 1.58) + 6.20$

$\\ x = 4.1776\frac{bl}{s}$

Thus, the 10th percentile is about x = 4.1776 bl/s and 90th percentile is about x = 8.2224 bl/s.

We can see the graphs below showing the cumulative probabilities for scores faster than 5, slower than 6, for the 10th percentile and the 90th percentile.