Question

1. the amount of snowfall in aspen, colorado for the winter averages 50 inches with a standard deviation of 15. x = snowfall amount on a randomly selected winter. find:

a. p(x > 75)
b. p(x < 35)
c. p(38 < x < 62)
d. p(60 < x < 70)
e. p(x = 50)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 50\) (mean) and \(\sigma=15\) (standard deviation).

• For \(x = 75\): \(z_1=\frac{75 - 50}{15}=\frac{25}{15}\approx1.67\)
• For \(x = 35\): \(z_2=\frac{35 - 50}{15}=\frac{-15}{15}=-1\)
• For \(x = 38\): \(z_3=\frac{38 - 50}{15}=\frac{-12}{15}=-0.8\)
• For \(x = 62\): \(z_4=\frac{62 - 50}{15}=\frac{12}{15}=0.8\)
• For \(x = 60\): \(z_5=\frac{60 - 50}{15}=\frac{10}{15}\approx0.67\)
• For \(x = 70\): \(z_6=\frac{70 - 50}{15}=\frac{20}{15}\approx1.33\)

Step2: Use the standard normal distribution table

• \(P(X>75)=P(Z > 1.67)\)

Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard - normal table \(P(Z\leq1.67) = 0.9525\), so \(P(Z>1.67)=1 - 0.9525=0.0475\)

• \(P(X < 35)=P(Z<-1)\)

From the standard - normal table \(P(Z<-1)=0.1587\)

• \(P(38 < X < 62)=P(-0.8<Z<0.8)\)

Since \(P(-a < Z < a)=2\Phi(a)-1\), and \(\Phi(0.8)=0.7881\), so \(P(-0.8 < Z < 0.8)=2\times0.7881-1=0.5762\)

• \(P(60 < X < 70)=P(0.67<Z<1.33)\)

\(P(0.67<Z<1.33)=P(Z < 1.33)-P(Z < 0.67)\)
From the standard - normal table \(P(Z < 1.33)=0.9082\) and \(P(Z < 0.67)=0.7486\)
\(P(0.67<Z<1.33)=0.9082 - 0.7486=0.1596\)

• For a continuous random variable \(X\) (since snowfall is a continuous variable), \(P(X = 50)=0\)

Answer:

A. \(0.0475\)
B. \(0.1587\)
C. \(0.5762\)
D. \(0.1596\)
E. \(0\)