Question

2. In a study of nerve cells, researchers measured the timing of spontaneous electrical discharges in a single organism. The interspike-time intervals were approximately normally distributed with a mean of 15.28 ms and a standard deviation of 0.75 ms. Suppose you measured a random sample of n = 4 interspike-time intervals from this population. a. What is the probability that all four times in your sample are between 15.0 and 16.0 ms? (3pt) b. What is the probability that the sample mean time is between 15.0 and 16.0 ms? (3pt) c. Comparing the probabilities you calculated in parts (a) and (b), which is larger? Even if we changed the mean and standard deviation of this population, briefly explain why this will always be the case. (2pt)

Answer 1

Answer:

Kindly check explanation

Step-by-step explanation:

Given that:

Mean (m) = 15.28

Standard deviation (σ) = 0.75

Sample size (n) = 4

What is the probability that all four times in your sample are between 15.0 and 16.0 ms?

x = 15

Zscore = (x - m) / (σ /√n)

Z = (15 - 15.28) / (0.75)

Z = −0.3733

P(Z < - 0.3733) = 0.35446

X = 16

Z = (16 - 15.28) / (0.75)

Z = 0.96

P(Z < 0.96) = 0.83147

0.83147 - 0.35446 = 0.47701

4 * 0.47701 = 1.90804

B.)

x = 15

Zscore = (x - m) / (σ /√n)

Z = (15 - 15.28) / (0.75/√4)

Z = −0.7467

P(Z < - 0.7467) = 0.22762

X = 16

Z = (16 - 15.28) / (0.75/√4)

Z = 1.92

P(Z < 1.92) = 0.97257

0.97257 - 0.22762 = 0.74495

C.) 1.9084 > 0.74495