Assume the heights of children are normally distributed with the mean and standard deviation you calculated in the last question. What percent of children have heights between 61.6 cm and 74.9 cm ?

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We assume that from the previous question the heights are normally distributed with mean  
$$
\mu = 68.6\text{ cm}
$$  
and standard deviation  
$$
\sigma = 4.9\text{ cm}.
$$

We wish to find the percentage of children with heights between 61.6 cm and 74.9 cm. This probability is given by

$$
P(61.6 \leq X \leq 74.9) = \Phi\left(\frac{74.9-\mu}{\sigma}\right) - \Phi\left(\frac{61.6-\mu}{\sigma}\right),
$$

where $$\Phi(z)$$ is the cumulative distribution function (CDF) of the standard normal distribution.

**Step 1. Standardize the lower bound**

The lower bound $$x = 61.6$$ corresponds to the $$z$$-score

$$
z_1 = \frac{61.6 - 68.6}{4.9} = \frac{-7}{4.9} \approx -1.4286.
$$

**Step 2. Standardize the upper bound**

The upper bound $$x = 74.9$$ corresponds to the $$z$$-score

$$
z_2 = \frac{74.9 - 68.6}{4.9} = \frac{6.3}{4.9} \approx 1.2857.
$$

**Step 3. Find the corresponding probabilities**

Using a standard normal table or calculator:

- For $$z_2 \approx 1.29$$, we have

$$
  \Phi(1.29) \approx 0.9015.
  $$

- For $$z_1 \approx -1.43$$, since symmetric properties give

$$
  \Phi(-1.43) = 1 - \Phi(1.43) \quad \text{and} \quad \Phi(1.43) \approx 0.9236,
  $$
  
  then

$$
  \Phi(-1.43) \approx 1 - 0.9236 = 0.0764.
  $$

**Step 4. Compute the probability between the two bounds**

Now the probability that a child’s height is between 61.6 cm and 74.9 cm is

$$
P(61.6 \leq X \leq 74.9) \approx 0.9015 - 0.0764 = 0.8251.
$$

Thus, approximately $$82.51\%$$ of children have heights between 61.6 cm and 74.9 cm.
Approximately 82.51% of children have heights between 61.6 cm and 74.9 cm.

Assume the heights of children are normally distributed with the mean and standard deviation you calculated in the last question. What percent of children have heights between 61.6 cm and 74.9 cm ?

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