H(x) =x3-7x2-x+7 helpp

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kim643

02.08.2020

Mathematics

The coach should start recruiting players with weight 269.55 pounds or more.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 225 pounds

Standard Deviation, σ = 43 pounds

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.15

$P( X x) = P( z \displaystyle\frac{x - 225}{43})=0.15$

$= 1 -P( z \leq \displaystyle\frac{x - 225}{43})=0.15$

$=P( z \leq \displaystyle\frac{x - 225}{43})=0.85$

Calculation the value from standard normal z table, we have,

P(z < 1.036) = 0.85

$\displaystyle\frac{x - 225}{43} = 1.036\\\\x = 269.548 \approx 269.55$

Thus, the coach should start recruiting players with weight 269.55 pounds or more.

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