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Please help me solve this as many as you can. Thank you. 1/ Doing research for...

Please help me solve this as many as you can. Thank you.

1/ Doing research for insurance rates, it is found that those aged 30 to 49 drive an average of 38.7 miles per day with a standard deviation of 6.7 miles. These distances are normally distributed. If a group of 60 drivers in that age group are randomly selected, what is the probability that the mean distance traveled each day is between 29.9 miles and 39.9 miles?

2/ A stock’s price fluctuations are approximately normally distributed with a mean of $29.51 and a standard deviation of $3.87. You decide to sell whenever the price reaches its highest 10% of values. What is the highest value you would still hold the stock?

Thank you again.

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Answer #1

1) P(29.9 < ar x < 39.9)

= P((29.9 - mu)/(sigma/sqrt n) < (ar x - mu)/(sigma/sqrt n) < (39.9 - mu)/(sigma/sqrt n))

= P((29.9 - 38.7)/(6.7//60) < Z < (39.9 - 38.7)/(6.7//60))

= P(-10.17 < Z < 1.39)

= P(Z < 1.39) - P(Z < -10)

= 0.9177 - 0

= 0.9177

2) P(X > x) = 0.10

or, P((X - mu)/sigma > (x - mu)/sigma) = 0.10

or, P(Z > (x - 29.51)/3.87) = 0.10

or, P(Z < (x - 29.51)/3.87) = 0.90

or, (x - 29.51)/3.87 = 1.28

or, x = 1.28 * 3.87 + 29.51

or, x = 34.46

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