Answer:
The probability that maximum safe load will be exceeded is 3.9%
Explanation:
We are given the following information in the question:
Mean, μ = 190 pounds
Standard Deviation, σ = 40 pounds
We are given that 50 skiers are randomly chosen from this normally distributed population, hence, by central limit theorem:
Mean, μ = 190 pounds
Standard Deviation,
![\sigma = \displaystyle(40)/(√(50))\text{ pounds}](https://img.qammunity.org/2020/formulas/mathematics/college/7hqzryjkxzkw41i7yl2vomj32nv5rey1fe.png)
Formula:
![z_(score) = \displaystyle(x-\mu)/(\sigma)](https://img.qammunity.org/2020/formulas/mathematics/college/5bpvqdbyqd8y38zhlcp80hz1p4ka5nivnl.png)
a) P(maximum safe load exceeded)
![P( x > (10000)/(50)) = P( z > \displaystyle((10000)/(50) - 190)/((40)/(√(50)))) = P(z > 1.767)](https://img.qammunity.org/2020/formulas/mathematics/college/ll9i4nj9wb27zibkgrj75d90eyjcufrr3x.png)
![= 1 - P(z \leq 1.767)](https://img.qammunity.org/2020/formulas/mathematics/college/p5a8nu3prgz570189i04mkzry86qmb9n4h.png)
Calculation the value from standard normal z table, we have,
![P(x > 10000) = 1 - 0.961 = 0.039= 3.9\%](https://img.qammunity.org/2020/formulas/mathematics/college/leqjtaflq0xjgeuu6w920qil7wglqxjoh6.png)
Hence, the probability that maximum safe load will be exceeded is 3.9%