Answer to Question #270548 in Statistics and Probability for casanova

Question

1. Baseball - A baseball franchise finds that the attendance at its home games is normally distributed, with a mean of 16,000 and a standard deviation of 4000.

a. What percent of the home games have an attendance between 12 ,000 and 20,000 people?

b. What percent of the home games have an attendance of fewer than 8000 people?

2. Find the area, to the nearest thousandth, of the indicated region of the standard normal distribution.

a. The region where z > 1.3

b. The region where z < 1.92

Answer 1

1. Let $X=$ the attendance at home games: $X\sim N(\mu, \sigma^2).$

Given $\mu=16000, \sigma=4000$

a. Use the 68–95–99.7 rule

$$16000-12000=4000=20000-16000=1\sigma$$

$$P(12000<X<20000)$$

$$=P(\mu-1\sigma<X<\mu+1\sigma)\approx0.6827$$

$68.27 \%$

Or

$$P(12000<X<20000)$$

$$=P(X<20000)-P(X\leq12000)$$

$$=P(Z<\dfrac{20000-16000}{4000})-P(Z\leq\dfrac{12000-16000}{4000})$$

$$=P(Z<1)-P(Z\leq-1)$$

$$\approx0.8413447-0.1586553$$

$$\approx0.6827$$

$68.27 \%$

b.

$$P(X<8000)=P(Z<\dfrac{8000-16000}{4000})$$

$$=P(Z<-2)\approx0.02275$$

$2.275 \%$

2.

a.

$$P(Z>1.3)=1-P(Z\leq 1.3)$$

$$\approx1-0.9031995\approx0.0968$$

b.

$$P(Z<1.92)\approx0.972571$$