Answer:
a) [tex]X\sim(246,1521)[/tex]
b) 0.1191
c) [tex]P_{70}=266.44[/tex]
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 246 feet
Standard Deviation, σ = 39 feet
We are given that the distribution of distance of fly balls is a bell shaped distribution that is a normal distribution.
a) Distribution of X
Let X be the distance in feet for a fly ball. Then,
[tex]X\sim (\mu, \sigma^2)\\X\sim(246,39^2)\\X\sim(246,1521)[/tex]
b) Probability that a randomly hit fly ball travels less than 200 feet.
[tex]P( x < 200) = P( z < \displaystyle\frac{200 - 246}{39}) = P(z < -1.1794)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 200) = 0.1191[/tex]
c) 70th percentile for the distribution of fly balls.
We have to find the value of x such that the probability is 0.7
[tex]P( X < x) = P( z < \displaystyle\frac{x - 246}{39})=0.7[/tex]
Calculation the value from standard normal z table, we have,
[tex]\displaystyle\frac{x - 246}{39} = 0.524\\\\x = 266.436\approx 266.44[/tex]
The 70th percentile for the distribution of fly ball is 266.44 feet.
