Question

suppose the age that children learn to walk is normally distributed with mean 12 months and standard deviation 2.1 month. 35 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 sim n(),
)
b. what is the distribution of ( overline{x} )? ( overline{x} sim n(),
)
c. what is the probability that one randomly selected person learned to walk when the person was between 11.5 and 12.4 months old?
d. for the 35 people, find the probability that the average age that they learned to walk is between 11.5 and 12.4 months old.
e. for part d), is the assumption that the distribution is normal necessary? ( \bigcirc \text{no} \bigcirc \text{yes} )
f. find the iqr for the average first time walking age for groups of 35 people.
( q1=) months
( q3=) months
iqr: months

Explanation:

Step1: Distribution of \(X\)

Given \(X\) (age children learn to walk) is normally distributed with mean \(\mu = 12\) and standard deviation \(\sigma=2.1\). So \(X\sim N(12,2.1)\)

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

For a sample of size \(n = 35\), the mean of the sampling distribution of the sample mean \(\bar{x}\) is \(\mu_{\bar{x}}=\mu = 12\), and the standard deviation of the sampling distribution (standard error) is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{2.1}{\sqrt{35}}\approx0.355\). So \(\bar{x}\sim N(12, 0.355)\)

Step3: Probability for \(X\)

First, find the \(z\) - scores. For \(x_1 = 11.5\), \(z_1=\frac{11.5 - 12}{2.1}\approx - 0.2381\). For \(x_2 = 12.4\), \(z_2=\frac{12.4 - 12}{2.1}\approx0.1905\). Then \(P(11.5<X<12.4)=\Phi(0.1905)-\Phi(-0.2381)\). Using the standard normal table or calculator, \(\Phi(0.1905)\approx0.5753\), \(\Phi(-0.2381)\approx0.4052\). So \(P(11.5<X<12.4)=0.5753 - 0.4052=0.1701\)

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

For \(\bar{x}_1 = 11.5\), \(z_1=\frac{11.5 - 12}{0.355}\approx - 1.4085\). For \(\bar{x}_2 = 12.4\), \(z_2=\frac{12.4 - 12}{0.355}\approx1.1268\). Then \(P(11.5<\bar{x}<12.4)=\Phi(1.1268)-\Phi(-1.4085)\). Using the standard normal table or calculator, \(\Phi(1.1268)\approx0.8697\), \(\Phi(-1.4085)\approx0.0799\). So \(P(11.5<\bar{x}<12.4)=0.8697 - 0.0799 = 0.7898\)

Step5: Normality assumption for part (d)

By the Central Limit Theorem, since \(n = 35\geq30\), the sampling distribution of \(\bar{x}\) is approximately normal regardless of the population distribution. So the answer is No.

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

The first quartile \(Q_1\): \(z=-0.6745\) (from standard normal distribution for \(25^{th}\) percentile). Using \(z=\frac{\bar{x}-12}{0.355}\), \(- 0.6745=\frac{\bar{x}-12}{0.355}\), \(\bar{x}=12-0.6745\times0.355\approx11.761\). The third quartile \(Q_3\): \(z = 0.6745\). Using \(z=\frac{\bar{x}-12}{0.355}\), \(\bar{x}=12 + 0.6745\times0.355\approx12.239\). The IQR \(=Q_3 - Q_1=12.239 - 11.761 = 0.478\)

Answer:

a. \(X\sim N(12,2.1)\)
b. \(\bar{x}\sim N(12,0.355)\)
c. \(0.1701\)
d. \(0.7898\)
e. No
f. \(Q1 = 11.761\) months, \(Q3 = 12.239\) months, \(IQR = 0.478\) months