Respuesta :
Answer:
a) 0.964
b) 0.500
c) 0.885
d) [tex]x \geq 32467.5[/tex]
e) 0.997
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 30000
Standard Deviation, σ = 1500
We are given that the distribution of attendance at stadium is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(attendance is greater than 27,300)
P(x > 27300)
[tex]P( x > 27300) = P( z > \displaystyle\frac{27300 - 30000}{1500}) = P(z > -1.8)[/tex]
[tex]= 1 - P(z \leq -1.8)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 27300) = 1 - 0.036 = 0.964 = 96.4\%[/tex]
b) P(attendance greater than or equal to 30000)
[tex]P(x > 30000) = P(z > \displaystyle\frac{30000-30000}{1500}) = P(z \geq 0)\\\\P( z \geq 0) = 1 - P(z \leq 0)[/tex]
Calculating the value from the standard normal table we have,
[tex]1 - 0.500 = 0.500 = 50\%[/tex]
c) P(attendance between 27000 and 32000)
[tex]P(27000 \leq x \leq 32000) = P(\displaystyle\frac{27000 - 30000}{1500} \leq z \leq \displaystyle\frac{32000-30000}{1500}) = P(-2 \leq z \leq 1.33)\\\\= P(z \leq 1.33) - P(z < -2)\\= 0.908 - 0.023 = 0.885 = 88.5\%[/tex]
[tex]P(27000 \leq x \leq 32000) = 88.5\%[/tex]
e) P(attendance less than 33000)
P(x < 33000)
[tex]P( x < 33000) = P( z < \displaystyle\frac{33000 - 30000}{1500}) = P(z < 2)[/tex]
Calculating the value from the standard normal table we have,
P(x < 33000) = 0.997 = 99.7%
d) We have to find x such that:
[tex]P( X > x) = P( z > \displaystyle\frac{x - 30000}{1500}) = 0.95[/tex]
Calculating the value from the standard normal table we have,
P(z = 1.645) = 0.95
Thus,
[tex]\displaystyle\frac{x - 30000}{1500} \geq 1.645\\\\x \geq 32467.5[/tex]
The attendance should be greater than or equal to 32467.5 to be in the top 5% of all games.
