10. The weight of adult bottlenose dolphins was found to follow a normal distribution with a mean of 550 pounds and a standard deviation of 50 pounds. a) What percentage of bottlenose dolphins weigh from 400 to 600 pounds? b) In a random sample of 9 adult bottlenose dolphins, what is the probability that exactly 5 of them are heavier than 560 pounds? (a) Let X be the weight of an adult bottlenose dolphin. 400 - 550 Then 600-550 50 <2< 50 = Þ(1) – Þ(−3) = 0.8413 – 0.0013 = 0.84 P(400 < X < 600) = P (b) The first step is to find the probability that a single dolphin weighs more than 560 pounds. P(X > 560) = P(Z > : P(Z > = P(−3 < Z < 1) 560-550 50 = P(Z > 0.2) = 1 – Þ(0.2) = 1 – 0.5793 = 0.4207 Let y be the number of dolphins which are heavier than 560 pounds out of the sample of 9. Then Y~Bin(9, 0.4207). The remainder of the solution can now be done two different ways. The first is to use the binomial distribution formula. P(Y = 5) = (²) (0.4207)5(0.5793)¹ = 0.1870 The second way to complete is to use the normal approximation to the binomial and find

A First Course in Probability (10th Edition)
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10. The weight of adult bottlenose dolphins was found to follow a normal distribution with a
mean of 550 pounds and a standard deviation of 50 pounds.
a) What percentage of bottlenose dolphins weigh from 400 to 600 pounds?
b) In a random sample of 9 adult bottlenose dolphins, what is the probability that exactly 5 of
them are heavier than 560 pounds?
(a) Let X be the weight of an adult bottlenose dolphin.
400 - 550
Then
600-550
50
<2<
50
= Þ(1) – Þ(−3) = 0.8413 – 0.0013 = 0.84
P(400 < X < 600) = P
(b) The first step is to find the probability that a single dolphin weighs more than 560
pounds.
P(X > 560) = P(Z >
: P(Z >
= P(−3 < Z < 1)
560-550
50
= P(Z > 0.2) = 1 – Þ(0.2) = 1 – 0.5793 = 0.4207
Let y be the number of dolphins which are heavier than 560 pounds out of the sample of
9. Then Y~Bin(9, 0.4207).
The remainder of the solution can now be done two different ways. The first is to use the
binomial distribution formula.
P(Y = 5) = (²) (0.4207)5(0.5793)¹ = 0.1870
The second way to complete is to use the normal approximation to the binomial and find
Transcribed Image Text:10. The weight of adult bottlenose dolphins was found to follow a normal distribution with a mean of 550 pounds and a standard deviation of 50 pounds. a) What percentage of bottlenose dolphins weigh from 400 to 600 pounds? b) In a random sample of 9 adult bottlenose dolphins, what is the probability that exactly 5 of them are heavier than 560 pounds? (a) Let X be the weight of an adult bottlenose dolphin. 400 - 550 Then 600-550 50 <2< 50 = Þ(1) – Þ(−3) = 0.8413 – 0.0013 = 0.84 P(400 < X < 600) = P (b) The first step is to find the probability that a single dolphin weighs more than 560 pounds. P(X > 560) = P(Z > : P(Z > = P(−3 < Z < 1) 560-550 50 = P(Z > 0.2) = 1 – Þ(0.2) = 1 – 0.5793 = 0.4207 Let y be the number of dolphins which are heavier than 560 pounds out of the sample of 9. Then Y~Bin(9, 0.4207). The remainder of the solution can now be done two different ways. The first is to use the binomial distribution formula. P(Y = 5) = (²) (0.4207)5(0.5793)¹ = 0.1870 The second way to complete is to use the normal approximation to the binomial and find
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