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We have been given that miss Roxanne is 25 years old and she puts 1800 dollars per quarter that returns 6% interest.
(a) We need to figure out how much will be in the account when she turns 65 years old. When she turns 65 years old, the number of years during which she made deposits would be 40. Since she made quarterly deposits. She made a total of 160 deposits. We can now figure out the final amount in the account using future value of annuity formula.
We have the values P=1800, r=6/4% = 1.5% = 0.015 and n=160.
Therefore, the amount in the account would be:
Therefore, miss Roxanne will be 1179415.39 dollars in her account when she turns 65 years old.
(b) In this part we need to figure out the total amount she deposited.
The total amount she deposited would be .
(c) We can find the interest earned by subtracting her contribution from the answer of part (a).
Interest earned =
Answer:
(a) 0.50928
(b) 0.857685.
Step-by-step explanation:
We are given that an engineer is going to redesign an ejection seat for an airplane. The new population of pilots has normally distributed weights with a mean of 155 lb and a standard deviation of 29.2 lb i.e.; = 160 lb and
= 27.5 lb
(A) We know that Z = ~ N(0,1)
Let X = randomly selected pilot
If a pilot is randomly selected, the probability that his weight is between 150 lb and 201 lb = P(150 < X < 201)
P(150 < X < 201) = P(X < 201) - P(X <= 150)
P(X < 201) = P( <
) = P(Z < 1.57) = 0.94179
P(X <= 150) = P( <
) = P(Z < -0.17) = 1 - P(Z < 0.17) = 1 - 0.56749
= 0.43251
Therefore, P(150 < X < 201) = 0.94179 - 0.43251 = 0.50928 .
(B) We know that for sampling mean distribution;
Z = ~ N(0,1)
If 39 different pilots are randomly selected, the probability that their mean weight is between 150 lb and 201 lb is given by P(150 < X bar < 210);
P(150 < X bar < 210) = P(X bar < 201) - P(X bar <= 150)
P(X bar < 201) = P( <
) = P(Z < 9.84) = 1 - P(Z >= 9.84)
= 0.999995
P(X bar <= 150) = P( <
) = P(Z < -1.07) = 1 - P(Z < 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(150 < X bar < 210) = 0.999995 - 0.14231 = 0.857685.
C) If the tolerance level is very high to accommodate an individual pilot then it should be appropriate ton consider the large sample i.e. Part B probability is more relevant in that case.