Question

The heights of children 2 years old are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches. pediatricians regularly measure the heights of toddlers to determine whether there is a problem. there may be a problem when a child is in the top or bottom 5% of heights. find the probability of these events.

Answer 1

Normal growth is a sign of good health. Monitoring for growth disturbances is fundamental to children's health care. Early detection and diagnosis of the causes of short stature allows management of underlying medical conditions, optimizing attainment of good health and normal adult height. Normal growth is a sign of good health. Monitoring for growth disturbances is fundamental to children's health care. Early detection and diagnosis of the causes of short stature allows management of underlying medical conditions, optimizing attainment of good health and normal adult height. Short stature with decreased weight-for-height ratio may indicate the presence of a chronic systemic disease.

Answer 2

(a). The height greater than 34.5 and less than 29.5 could be a problem.

(b). The probability that a height greater than 36 is $\boxed{0.0038}$ and the height less than 34 is $\boxed{0.9082}.$

(c). The probability that the height of child is between 30 and 33 inches tall is $\boxed{0.6568}.$

Further Explanation:

The Z score of the standard normal distribution can be obtained as,

${\text{Z}} = \dfrac{{X - \mu }}{\sigma }$

Given:

The mean of is $\boxed{32}.$

The standard deviation of the waiting time is $\boxed{1.5}.$

Explanation:

Part (a),

The probability is less than $5\%.$

$\begin{aligned}P\left( {Z < z} \right) &= 0.05\\\frac{{X - \mu }}{\sigma } &= 1.645\\\frac{{X - 32}}{{1.5}} &= 1.645\\X - 32 &= 1.645 \times 1.5\\X &= 32 + 2.47\\X &= 34.47\\\end{aligned}$

Part(b),

The probability that a height greater than 36 can be obtained as follows,

$\begin{aligned}P\left( {X > 36} \right) &= P\left( {Z > \frac{{36 - 32}}{{1.5}}} \right)\\&= P\left( {Z > 2.67} \right)\\&= 1 - P\left( {Z < 2.67} \right)\\&= 0.0038\\\end{aligned}$

The probability that height less than 34 can be obtained as follows,

$\begin{aligned}P\left( {X < 34} \right) &= P\left( {Z < \frac{{34 - 32}}{{1.5}}} \right)\\&= P\left( {Z < 1.33} \right)\\&= 0.9082\\\end{aligned}$

Part (c),

$\begin{aligned}P\left( {30 < X < 33} \right) &= P\left( {\frac{{30 - 32}}{{1.5}} < Z < \frac{{33 - 32}}{{1.5}}} \right)\\&= P\left( { - 1.33 < Z < 0.67} \right)\\&=0.6568\\\end{aligned}$

Learn more:

1. Learn more about normal distribution brainly.com/question/12698949
2. Learn more about standard normal distribution brainly.com/question/13006989
3. Learn more about confidence interval of meanhttps://brainly.com/question/12986589

Answer details:

Grade: College

Subject: Statistics

Chapter:Normal distribution

Keywords: heights of children, 2 years old, heights of toddlers, normally distributed, mean, mean of 32 inches, standard deviation of 1.5 inches, pediatricians, probability, events, 5% of heights.