Question

Question 1 In a learning experiment, untrained mice are placed in a maze and the time required for each mouse to exit the maze is recorded. The average time for untrained mice to exit the maze is μ = 50 seconds and the standard deviation of their times is ơ-16 seconds. Sixty four randomly selected untrained mice are placed in the maze and the time necessary to exit the maze is recorded for each. a) Completely describe the sampling distribution of the sample mean. b) What is the probability that the sample mean differs from the population mean by more than 3? c) If one mouse is selected at random, what is the probability that the time necessary to exit the maze differs from the population mean by more than 3?

Answer 1

a) $\mu_{\bar x}$ = 50

   $\sigma_{\bar x}$ = $\sigma/\sqrt n$

        = 16/$\sqrt 64$ = 2

b) P(47 < $\bar x$ < 53)

= P((47 - $\mu$)/($\sigma/\sqrt n$) < ($\bar x$ - $\mu$)/($\sigma/\sqrt n$) < (53 - $\mu$)/($\sigma/\sqrt n$))

= P((47 - 50)/2 < Z < (53 - 50)/2)

= P(-1.5 < Z < 1.5)

= P(Z < 1.5) - P(Z < -1.5)

= 0.9332 - 0.0668

= 0.8664

c) P(47 < X < 53)

= P((47 - $\mu$)/$\sigma$ < (X - $\mu$)/$\sigma$ < (53 - $\mu$)/$\sigma$)

= P((47 - 50)/16 < Z < (53 - 50)/16)

= P(-0.19 < Z < 0.19)

= P(Z < 0.19) - P(Z < -0.19)

= 0.5753 - 0.4247

= 0.1506

d) part b) There is no need for any assumption about population. Since the sample size is large(n > 30), so according to the Central limit theorem the sampling distribution of the sample mean will be approximately normally distributed.

part c) Assume that the population is normally distributed. Since the sample size is not large, so we cannot apply the Central limit theorem.

e) P($\bar x$ < 85)

= P(($\bar x$ - $\mu$)/($\sigma/\sqrt n$) < (85 - $\mu$)/($\sigma/\sqrt n$))

= P( Z < (85 - 50)/2)

= P(Z < 17.5)

= 1

Since the probability is greater than 0.95, so it is unusual.

f) We cannot support the claim that the average time for untrained mice to exit the maze is equal to 50 seconds.