Question

heights of 10 year old children, regardless of sex, closely follow a normal distribution with mean 55 inches and standard deviation 6.1 inches. round answers to 4 decimal places.
a) what is the probability that a randomly chosen 10 year old child is less than 47.2 inches?
b) what is the probability that a randomly chosen 10 year old child is more than 57.7 inches?
c) what proportion of 10 year old children are between 51.6 and 60.2 inches tall?
d) 75% of all 10 year old child are less than inches.

Explanation:

Step1: Calculate z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the value from the data set, \(\mu\) is the mean and \(\sigma\) is the standard deviation.

Step2: Solve part (a)

Given \(\mu = 55\), \(\sigma=6.1\), \(x = 47.2\)
\(z=\frac{47.2 - 55}{6.1}=\frac{- 7.8}{6.1}\approx - 1.28\)
Using the standard normal table \(P(Z\lt - 1.28)=0.1003\)

Step3: Solve part (b)

Given \(x = 57.7\)
\(z=\frac{57.7-55}{6.1}=\frac{2.7}{6.1}\approx0.44\)
\(P(Z\gt0.44)=1 - P(Z\leqslant0.44)\)
From the standard normal table \(P(Z\leqslant0.44) = 0.6700\)
\(P(Z\gt0.44)=1 - 0.6700=0.3300\)

Step4: Solve part (c)

For \(x_1 = 51.6\), \(z_1=\frac{51.6 - 55}{6.1}=\frac{-3.4}{6.1}\approx - 0.56\)
For \(x_2 = 60.2\), \(z_2=\frac{60.2-55}{6.1}=\frac{5.2}{6.1}\approx0.85\)
\(P(-0.56\lt Z\lt0.85)=P(Z\lt0.85)-P(Z\lt - 0.56)\)
From the standard normal table \(P(Z\lt0.85)=0.8023\), \(P(Z\lt - 0.56)=0.2877\)
\(P(-0.56\lt Z\lt0.85)=0.8023 - 0.2877=0.5146\)

Step5: Solve part (d)

We need to find \(x\) such that \(P(Z\lt z)=0.75\)
Looking up in the standard normal table, the \(z\) - score corresponding to a probability of \(0.75\) is \(z = 0.67\)
Using \(z=\frac{x-\mu}{\sigma}\), we have \(0.67=\frac{x - 55}{6.1}\)
\(x=55+0.67\times6.1=55 + 4.087=59.087\approx59.09\)

Answer:

a) \(0.1003\)
b) \(0.3300\)
c) \(0.5146\)
d) \(59.09\)