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.
1) P(29.9 < < 39.9)
= P((29.9 - )/() < ( - )/() < (39.9 - )/())
= P((29.9 - 38.7)/(6.7/) < Z < (39.9 - 38.7)/(6.7/))
= 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 - )/ > (x - )/) = 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
Please help me solve this as many as you can. Thank you. 1/ Doing research for...
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?
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