Statistics. Please check (b) - (c). Need help with (a).?

Let x be a random variable that represents the incubation time for Allen hummingbird eggs. The x distribution has a mean of µ = 16 days. Let us assume that the standard deviation is approximately σ = 2 days. The distribution of x values is more or less mound-shaped and symmetrical but not necessarily normal. Suppose that we have n = 30 eggs in an incubator. Let x be the average incubation time for these eggs.





(a) What can we say about the probability distribution of x? Is it approximately normal? What are the mean and standard deviation?





(b) What is the probability that x is between 16 and 17 days?


µ = 16 σ = 2 n = 30


P(16 %26lt; x-bar %26lt; 17)


= P(xBar %26lt; 17) - P(xBar %26lt; 16)


17 – 16 = -1


-1 / (2 / √30) = 2.74


16 – 16 = 0


0 / (2 / √30) = 0


0.9969 – 0.5000 = 0.4969


There is a 50% chance that the eggs will hatch between 16 and 17 days.





(c) What is the probability that x is less than 15 days?


P(Xbar %26lt; 15)


= 15 – 16 = -1


-1 / (2 / √30) = -2.74


= 0.0031

Statistics. Please check (b) - (c). Need help with (a).?
(a)





we can't say anything about the distribution of x itself other than what has been said in the problem statement. The central limit theorem tells us that the mean of any sample from any distribution with finite mean and variance will be approximately normal for a sufficient sample size. Since the distribution of x is mound shaped and the sample size is at the empirical magic value of 30 it is very plausible the mean will behave as a normal random variable.





The Central limit theorem is:


Let X1, X2, ... , Xn be a simple random sample from a population with mean μ and variance σ².





Let Xbar be the sample mean = 1/n * ∑Xi


Let Sn be the sum of sample observations: Sn = ∑Xi





then, if n is sufficiently large:





Xbar has the normal distribution with mean μ and variance σ² / n


Xbar ~ Normal(μ , σ² / n)





Sn has the normal distribution with mean nμ and variance nσ²


Sn ~ Normal(nμ , nσ²)





The great thing is that it does not matter what the under lying distribution is, the central limit theorem holds. It was proven by Markov using continuing fractions.





if the sample comes from a uniform distribution the sufficient sample size is as small as 12


if the sample comes from an exponential distribution the sufficient sample size could be several hundred to several thousand.





if the data comes from a normal distribution to start with then any sample size is sufficient.


for n %26lt; 30, if the sample is from a normal distribution we use the Student t statistic to estimate the distribution. We do this because the Student t takes into account the uncertainty in the estimate for the standard deviation.


if we now the population standard deviation then we can use the z statistic from the beginning.


the value of 30 was empirically defined because at around that sample size, the quantiles of the student t are very close the quantiles of the standard normal.





== -- == -- == -- == -- == -- == -- ==





(b)





In this question we have


Xbar ~ Normal( μ = 16 , σ² = 4 / 30 )


Xbar ~ Normal( μ = 16 , σ² = 0.1333333 )


Xbar ~ Normal( μ = 16 , σ = 2 / sqrt( 30 ) )


Xbar ~ Normal( μ = 16 , σ = 0.3651484 )





Find P( 16 %26lt; Xbar %26lt; 17 )


= P( ( 16 - 16 ) / 0.3651484 %26lt; ( Xbar - μ ) / σ %26lt; ( 17 - 16 ) / 0.3651484 )


= P( 0 %26lt; Z %26lt; 2.738613 )


= P( Z %26lt; 2.738613 ) - P( Z %26lt; 0 )


= 0.996915 - 0.5


= 0.4969151





== -- == -- == -- == -- == -- == -- == -- ==





(c)





Find P( Xbar %26lt; 15 )


P( ( Xbar - μ ) / σ %26lt; ( 15 - 16 ) / 0.3651484 )


= P( Z %26lt; -2.738613 )


= 0.003084950

our song