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Can anyone help me with this problem.

In a large number group of children $55\%$ are under $60$cm height and $40\%$ are between $60$cm and $65$cm. Assuming a normal distribution find the mean and standard deviation of height.

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    Do you know how to translate the information into the probabilities of a normal distribution? For instance the question is giving you the values of $P(X<60)$ and $P(60$X=N(\mu,\sigma)$ with $\mu$ and $\sigma$ unknown. – 2017-01-03
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    My frnd could u plz solve it. – 2017-01-03
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    Just write the two equations above in terms of $\mu$ and $\sigma$ and solve them... This link might help into which expressions do those equations look like: https://en.wikipedia.org/wiki/Normal_distribution#Quantile_function – 2017-01-03

1 Answers 1

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First, it is helpful to convert this into mathematical expressions.

We have:

$$P(X<60)=0.55$$ $$P(60

Where $X$ denotes the height. We notice that: $$P(X<65)=0.55+0.40=0.95$$ Hence, in words, $95\text{%}$ of children are under $65~\text{cm}$ of height.

We use the formula for the $z$-score to solve simultaneously for $P(X<60)=0.55$ and $P(X<65)=0.55+0.40=0.95$:

$$z=\frac{X-\mu}{\sigma} \tag{1}$$

Where $\mu$ is the mean and $\sigma$ is the standard deviation.

Keep in mind that $Z\sim N(0,1)$.

Hence, to find the $z$-scores, find the inverse normal of $0.55$ and $0.95$ under the distribution of $Z$.

Hence, we have from equation $(1)$:

$$\Phi^{-1}(0.55)=\frac{60-\mu}{\sigma}$$

$$\Phi^{-1}(0.95)=\frac{65-\mu}{\sigma}$$

$$0.125661\sigma+\mu=60$$ $$1.64485\sigma+\mu=65$$

Solving simultaneously, we obtain:

$$\mu \approx 59.5864$$ $$\sigma \approx 3.29123$$