Question

Need help on all parts please and thank you!

19. (0/3.33 Points] DETAILS PREVIOUS ANSWERS ILLOWSKYINTROSTAT1 7.HW.076. The attention span of a two-year-old is exponentially distributed with a mean of about 9 minutes. Suppose we randomly survey 60 two-year-olds. Parta) In words, define the random variable X. the attention span, in minutes, of all children O the attention span, in minutes, of a two-year-old child O the ages of the children with short attention spans O the number of two-year-old children X Part (0) Give the distribution of X X- Part (0) In words, define the random variable X the average age of children with short attention spans O the average attention span, in minutos, for sample of 60 children O the average attention span, in minutes, for all children the average number of two-year-old children Part (di Give the distribution of X. (Round your standard deviation to two decimal places.) X- Partial

Answer 1

The attention span of a two year old is exponentially distributed with a mean of 9 minutes.

Let X minutes be the random variable indicating the attention span of a two year old (Note that value of X is different for each two year old child).

X is exponentially distributed wit parameter

= 0.1111 mean 9

The cdf of X is

$\begin{align*} P(X<x)&=1-e^{\lambda x}=1-e^{\frac{1}{9}x},\quad x>0 \end{align*}$

part a) The random variable X is

ans:

the attention span, in minutes, of a two-year-old child

part b)

The distribution of X is

ans:

$X\sim \text{Exponential}(1/9)$ expressed as a fraction

or

expressed as a decimal

X Exponential(0.1111

part c) We have surveyed a sample of n=60 two-year-olds

Let be the sample mean attention span of a randomly selected sample of size n=60 two-year-olds.

X

ans:

the average attention span, in minutes, for sample of 60 children

part d) X is exponentially distributed with parameter $\lambda=\frac{1}{9}=0.1111$

The expected value (or the mean) of X is (using the results for Exponential distribution)

$\mu=E(X)=\frac{1}{\lambda}=\frac{1}{\frac{1}{9}}=9$

The standard deviation of X is (using the results for Exponential distribution)

$\sigma=\frac{1}{\lambda}=\frac{1}{1/9}=9$

The sample size n=60 is greater than 30. Using the central limit theorem we can say that
X
is normally distributed with mean $\mu_{\bar{X}}=\mu=9$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{9}{\sqrt{60}}=1.1619$

ans:

$\bar{X}\sim N(9,1.16)$

Note: The above assuming the second term is standard deviation. However, the standard notation of a normal distribution is

$\bar{X}\sim N(\mu,\sigma^2)=N(9,1.16^2)$

part e) the probability that an individual attention span is less than 10 minutes is

the probability that the average attention span for the 60 children is less than 10 minutes is $P(\bar{X}<10)$

We can see that
X
has a smaller standard deviation (1.16) compared to X (the standard deviation is 9). That is
X
has less spread than X.

Hence the Z score for 10 in P(X<10) is higher for X compared to 10 in $P(\bar{X}<10)$

That means 10 is farther to the right of mean of the distribution of , compared to that of X.

X

Hence the probability that the average attention span for the 60 children is less than 10 minutes is higher than the probability that an individual attention span is less than 10 minutes

ans: Which is higher?

the probability that the average attention span for the 60 children is less than 10 minutes

part f) the probability that an individual attention span is less than 10 minutes is (using the cdf of X)

$\begin{align*} P(X<10)=1-e^{\frac{1}{9}\times 10}=0.6708 \end{align*}$

ans:

the probability that an individual attention span is less than 10 minutes is 0.6708

the probability that the average attention span for the 60 children is less than 10 minutes is (using the normal distribution)

$\begin{align*} P(\bar{X}<10)&=P\left(\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}<\frac{10-\mu_{\bar{X}}}{\sigma_{\bar{X}}} \right )\quad\text{get the z score of 10}\\ &=P\left(Z<\frac{10-9}{1.16} \right )\\ &=P(Z<0.86)\\ &=0.8051\quad\text{using the standard normal table for z=0.86} \end{align*}$

ans:

the probability that the average attention span for the 60 children is less than 10 minutes is 0.8051

part g)

ans:

According to the Central Limit Theorem, as the number of samples increases, the sample means tend to follow a normal distribution

Note: in the above statement , rather than "...as the the number of samples increases...", the correct terminology is "...as the sample size increases..."