Question

unit 2: the normal distribution
l4 practice - the standard normal curve
name:

1. scores on the wechsler adult intelligence scale, a standard iq test, are approximately normal for the age

group of 20 to 34. the mean score is 110 with a standard deviation of 25. what percent of this age group has an
iq below 100?
a) write the inequality using the raw score.
b) find the z - score and rewrite the inequality.
c) draw the normal standardized curve.
d) what are you entering into your calculator?
lower:
upper:
mean ($\mu$):
standard deviation ($\sigma$):
e) % of adults from 20 to 34 have an iq less than 100.

1. the distribution of blood cholesterol levels in 14 - year - old boys is roughly normal; the mean is 165

milligrams of cholesterol per deciliter of blood and the standard deviation is 30. what proportion of 14 - year - ol
boys have a blood cholesterol level over 120 mg/dl?
a) write the inequality.
b) find the z - score and rewrite the inequality.
c) draw the normal standardized curve.
d) what are you entering into your calculator?
lower:
upper:
mean ($\mu$):
standard deviation ($\sigma$):
e) % of 14 - year - old boys have a blood cholesterol level over 120 mg/dl.

Explanation:

Step1: Calculate the z - score for IQ problem

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). For the IQ problem, \(x = 100\), \(\mu=110\), \(\sigma = 25\).

$$z=\frac{100 - 110}{25}=\frac{- 10}{25}=-0.4$$

Step2: Use the normal distribution function on the calculator

For a normal distribution \(N(\mu,\sigma)\), to find \(P(X<100)\) where \(X\sim N(110,25^{2})\), on a calculator (e.g., TI - 84), we use the normalcdf function. The lower bound is \(-\infty\) (we can use a very small number like \(-10^{9}\)), the upper bound is \(100\), \(\mu = 110\), \(\sigma=25\).

$$P(X < 100)=\text{normalcdf}(-10^{9},100,110,25)$$

Using the standard normal table or calculator, \(P(Z<-0.4)=0.3446\)

Step3: Calculate the z - score for cholesterol problem

For the cholesterol problem, \(x = 120\), \(\mu = 165\), \(\sigma=30\)

$$z=\frac{120 - 165}{30}=\frac{-45}{30}=-1.5$$

Step4: Use the normal distribution function on the calculator

To find \(P(X>120)\) where \(X\sim N(165,30^{2})\), we know that \(P(X>120)=1 - P(X\leq120)\). Using the normalcdf function, the lower bound is \(120\), the upper bound is \(10^{9}\), \(\mu = 165\), \(\sigma = 30\)

$$P(X>120)=\text{normalcdf}(120,10^{9},165,30)$$

Using the standard normal table or calculator, \(P(Z>-1.5)=1 - P(Z\leq - 1.5)=1-0.0668 = 0.9332\)

Answer:

1)
a) \(X<100\)
b) \(z=-0.4\), \(Z < - 0.4\)
d) Lower: \(-10^{9}\), Upper: \(100\), Mean \((\mu)\): \(110\), Standard Deviation \((\sigma)\): \(25\)
e) \(34.46\)
2)
a) \(X > 120\)
b) \(z=-1.5\), \(Z>-1.5\)
d) Lower: \(120\), Upper: \(10^{9}\), Mean \((\mu)\): \(165\), Standard Deviation \((\sigma)\): \(30\)
e) \(93.32\)