A gondola (cable car) at a ski area holds 50 people. Its maximum safe load is 10000 pounds. A populationof skiers has a distribution of weights with mean 190 pounds and standard deviation 40 pounds. If 50 skiersare randomly chosen from this population, and take the gondola, what is the probability its maximum safeload will be exceeded? (Hint: This question is about a sum, but a sum ofnitems is their mean timesn, sothis question can be understood as a question about a mean.)

Answer:The probability that maximum safe load will be exceeded is 3.9%Step-by-step explanation:We are given the following information in the question:Mean, μ = 190 poundsStandard Deviation, σ = 40 poundsWe are given that 50 skiers are randomly chosen from this normally distributed population, hence, by central limit theorem:Mean, μ = 190 poundsStandard Deviation,[tex]\sigma = \displaystyle\frac{40}{\sqrt{50}}\text{ pounds}[/tex]Formula:[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]a) P(maximum safe load exceeded)[tex]P( x > \frac{10000}{50}) = P( z > \displaystyle\frac{\frac{10000}{50} - 190}{\frac{40}{\sqrt{50}}}) = P(z > 1.767)[/tex][tex]= 1 - P(z \leq 1.767)[/tex]Calculation the value from standard normal z table, we have,  [tex]P(x > 10000) = 1 - 0.961 = 0.039= 3.9\%[/tex]Hence, the probability that maximum safe load will be exceeded is 3.9%