Question

suppose the age that children learn to walk is normally distributed with mean 11 months and standard deviation 1.1 month. 12 randomly selected people were asked what age they learned to walk. round all answers to 4 decimal places where possible.
a. what is the distribution of x? x ~ n(, )
b. what is the distribution of \\( \overline{x} \\)? \\( \overline{x} \\) ~ n(, )
c. what is the probability that one randomly selected person learned to walk when the person was between 10 and 12.5 months old?
d. for the 12 people, find the probability that the average age that they learned to walk is between 10 and 12.5 months old.
e. for part d), is the assumption that the distribution is normal necessary? yes no
f. find the iqr for the average first time walking age for groups of 12 people.
q1 = months
q3 = months
iqr: months

Explanation:

Step1: Distribution of \(X\)

Given \(X\) (age children learn to walk) is normally distributed with mean \(\mu = 11\) and standard deviation \(\sigma=1.1\). So \(X\sim N(11,1.1^{2})\)

Step2: Distribution of \(\bar{x}\)

For a sample of size \(n = 12\), the mean of the sampling distribution of the sample mean \(\bar{x}\) is \(\mu_{\bar{x}}=\mu = 11\) and the standard deviation is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{1.1}{\sqrt{12}}\approx0.3175\). So \(\bar{x}\sim N(11,0.3175^{2})\)

Step3: Probability for one person (\(X\))

First, find \(z\)-scores. For \(x = 10\), \(z_{1}=\frac{10 - 11}{1.1}\approx - 0.9091\). For \(x = 12.5\), \(z_{2}=\frac{12.5-11}{1.1}\approx1.3636\). Then \(P(10<X<12.5)=\Phi(1.3636)-\Phi(-0.9091)\)
Using standard normal table or calculator: \(\Phi(1.3636)\approx0.9131\), \(\Phi(-0.9091)\approx0.1807\). So \(P(10 < X < 12.5)=0.9131-0.1807 = 0.7324\)

Step4: Probability for sample mean (\(\bar{x}\))

For \(\bar{x}=10\), \(z_{1}=\frac{10 - 11}{0.3175}\approx - 3.15\). For \(\bar{x}=12.5\), \(z_{2}=\frac{12.5 - 11}{0.3175}\approx4.72\). Then \(P(10<\bar{x}<12.5)=\Phi(4.72)-\Phi(-3.15)\)
Using standard normal table or calculator: \(\Phi(4.72)\approx1\), \(\Phi(-3.15)\approx0.0008\). So \(P(10 < \bar{x}<12.5)=1 - 0.0008=0.9992\)

Step5: Assumption for part d)

Since the original population (\(X\)) is normally distributed, the assumption that the distribution is normal is not necessary for the sampling distribution of \(\bar{x}\) (by Central Limit Theorem, but here original is normal so even for small \(n = 12\) it's exact). But the answer is \(No\) (because for sampling distribution of \(\bar{x}\) when original is normal, it's always normal regardless of sample size)

Step6: IQR for \(\bar{x}\)

First, find \(z\)-scores for \(Q1\) (\(25^{th}\) percentile) and \(Q3\) (\(75^{th}\) percentile). \(z_{Q1}=- 0.6745\), \(z_{Q3}=0.6745\)
For \(Q1\): \(\bar{x}_{Q1}=\mu_{\bar{x}}+z_{Q1}\sigma_{\bar{x}}=11+(-0.6745)\times0.3175\approx10.786\)
For \(Q3\): \(\bar{x}_{Q3}=\mu_{\bar{x}}+z_{Q3}\sigma_{\bar{x}}=11 + 0.6745\times0.3175\approx11.214\)
\(IQR=\bar{x}_{Q3}-\bar{x}_{Q1}=11.214 - 10.786=0.428\)

Answer:

a. \(11\), \(1.1\)
b. \(11\), \(0.3175\)
c. \(0.7324\)
d. \(0.9992\)
e. \(No\)
f. \(Q1 = 10.786\), \(Q3 = 11.214\), \(IQR = 0.428\)