Question #270548

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




1
Expert's answer
2021-11-24T10:32:53-0500

1. Let X=X= the attendance at home games: XN(μ,σ2).X\sim N(\mu, \sigma^2).

Given μ=16000,σ=4000\mu=16000, \sigma=4000

a. Use the 68–95–99.7 rule


1600012000=4000=2000016000=1σ16000-12000=4000=20000-16000=1\sigma


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

=P(μ1σ<X<μ+1σ)0.6827=P(\mu-1\sigma<X<\mu+1\sigma)\approx0.6827

68.27%68.27 \%

Or


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

=P(X<20000)P(X12000)=P(X<20000)-P(X\leq12000)


=P(Z<20000160004000)P(Z12000160004000)=P(Z<\dfrac{20000-16000}{4000})-P(Z\leq\dfrac{12000-16000}{4000})

=P(Z<1)P(Z1)=P(Z<1)-P(Z\leq-1)

0.84134470.1586553\approx0.8413447-0.1586553

0.6827\approx0.6827

68.27%68.27 \%


b.

P(X<8000)=P(Z<8000160004000)P(X<8000)=P(Z<\dfrac{8000-16000}{4000})

=P(Z<2)0.02275=P(Z<-2)\approx0.02275

2.275%2.275 \%


2.

a.

P(Z>1.3)=1P(Z1.3)P(Z>1.3)=1-P(Z\leq 1.3)

10.90319950.0968\approx1-0.9031995\approx0.0968

b.


P(Z<1.92)0.972571P(Z<1.92)\approx0.972571



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