Question

In the nerve-cell activity of a certain individual fly, the time intervals between "spike" discharges follow approximately a normal distribution with mean \(15.6 \mathrm{~ms}\) and standard deviation \(0.4 \mathrm{~ms}\) (as in Example 4.1.3). Let \(Y\) denote a randomly selected interspike interval. Find (a) \(\operatorname{Pr}\\{Y>15\\}\) (b) \(\operatorname{Pr}\\{Y>16.5\\}\) (c) \(\operatorname{Pr}\\{15

Short answer

The probabilities for each interval are obtained using the corresponding Z-scores and the standard normal distribution table or calculator: (a) \(\operatorname{Pr}\{Y>15\}\), (b) \(\operatorname{Pr}\{Y>16.5\}\), (c) \(\operatorname{Pr}\{15<Y<16.5\}\), (d) \(\operatorname{Pr}\{15<Y<15.5\}\).

Step-by-step solution

Standardize the random variable Y

To find the probabilities involving the random variable Y which follows a normal distribution with mean \(\mu = 15.6 \) ms and standard deviation \(\sigma = 0.4 \) ms, we first convert Y to the standard normal variable Z using the formula \( Z = \frac{Y - \mu}{\sigma} \).

Find the Z-score for \(Y > 15\) ms

Calculate the Z-score corresponding to \(Y = 15 \) using the formula from Step 1. \[Z = \frac{15 - 15.6}{0.4} = -1.5\] This Z-score helps us find the probability for \(Y > 15\) ms.

Calculate \(\operatorname{Pr}\{Y>15\}\)

Use the Z-score obtained in Step 2 to find the area to the right of \(Z = -1.5\) on the standard normal distribution curve, which gives us \(\operatorname{Pr}\{Y>15\}\). This area can be found using a standard normal distribution table or a calculator that can compute normal probabilities.

Find the Z-score for \(Y > 16.5\) ms

Compute the Z-score for \(Y = 16.5\) ms. \[Z = \frac{16.5 - 15.6}{0.4} = 2.25\] This Z-score is used to determine \(\operatorname{Pr}\{Y>16.5\}\).

Calculate \(\operatorname{Pr}\{Y>16.5\}\)

Determine the area to the right of \(Z = 2.25\) to find \(\operatorname{Pr}\{Y>16.5\}\) using a standard normal distribution table or a probability calculator.

Find cumulative probabilities for \(15

We already have the Z-scores for \(Y = 15\) ms (\(Z = -1.5\)) and \(Y = 16.5\) ms (\(Z = 2.25\)) from previous steps. Find the cumulative probabilities for these Z-scores and subtract them to get \(\operatorname{Pr}\{15<Y<16.5\}\).

Find cumulative probabilities for \(15

First, calculate the Z-score for \(Y = 15.5\) ms. \[Z = \frac{15.5 - 15.6}{0.4} = -0.25\] Then, find the cumulative probabilities for \(Z = -1.5\) and \(Z = -0.25\) and subtract them to get \(\operatorname{Pr}\{15<Y<15.5\}\).

Key concepts

Probability Density Function

Understanding the probability density function (PDF) is crucial when dealing with continuous random variables like the time intervals between nerve-cell 'spike' discharges. In the context of our exercise, the PDF represents the distribution of these time intervals. A key concept to remember is that the area under the PDF curve represents probability. For a normal distribution, the curve is symmetrical and bell-shaped, centered around the mean. The total area under the curve adds up to 1, which corresponds to the certainty that a randomly selected interval will fall somewhere on the time axis.

In the example, the mean interspike interval is 15.6 ms. The probability of observing a specific interval, say exactly 15 ms, is technically zero since there is an infinite range of possible values. Instead, we calculate the probability over an interval of time, such as \(Y > 15\) ms, by finding the area under the PDF curve to the right of 15 ms.

Standard Deviation

Standard deviation (\(\sigma\)) measures the spread of a set of observations. In the case of our normally distributed nerve-cell 'spike' discharges, the standard deviation of 0.4 ms tells us how much the interspike intervals deviate from the mean (15.6 ms) on average.

A smaller standard deviation indicates that the data points are closer to the mean, suggesting tightly grouped intervals; conversely, a larger standard deviation signifies a wider spread, implying more variability in interspike duration. This concept is fundamental for transforming our random variable \(Y\) into a standard normal variable (\(Z-score\)) to easily find probabilities using standardized tables or tools.

Z-score

The Z-score is a type of standard score that indicates how many standard deviations an element is from the mean. For a normal distribution, converting a raw score into a Z-score allows you to determine where that value lies in the distribution. This standardization transforms any normal distribution into the standard normal distribution, which has a mean of 0 and a standard deviation of 1.

In the exercise, we calculate the Z-scores (like \(Z = -1.5\) for \(Y = 15\) ms) to find out how likely certain intervals are when compared to the average interval time. With these Z-scores, we can easily refer to standard normal probability tables to find the probabilities associated with different ranges of \(Y\).

Cumulative Probability

Cumulative probability is the likelihood that a variable will take a value less than or equal to a specific value. In terms of the normal distribution, it is represented graphically by the area under the PDF curve to the left of the specified value. This concept is vital for answering questions that ask for the probability within a range, as in parts (c) and (d) of the problem.

To find cumulative probabilities, we use the Z-scores calculated from the raw data. For \(15 < Y < 16.5\), we determine the cumulative probabilities for \(Z = -1.5\) and \(Z = 2.25\) and subtract the smaller from the larger. The result gives the total area between these two Z-scores, which is the probability of \(Y\) falling within the specified range.