Opisthotrochopodus n sp. is a polychaete worm that inhabits deep sea hydrothermal vents along the Mid-Atlantic Ridge. According to an article by Van Dover et al. in Marine Ecology Progress Series (1999, vol 181 pp 201-214), the lengths of female polychaete worms are normally distributed with mean 6.1 mm and standard deviation 1.3 mm. Let X denote length for female polychaete worms. Find the percentage of female polychaete worms that have lengths between 7 mm and 8 mm.

6.9: Given that X is normally distributed random variable with a mean of 60 and a standard deviation of 10, find the following probabilities:
Hint: The expression ‘P (X>60)’ stands for the area to the right of the normal variable X.
a. P (X>60) b. P (60<X<72)

c. P (57<X<83) d. P (65<X<82)

e. P (38<X<78) f. P (X<38)

I did the first part by entering the normaladf in my calculator as a result I got .172.

I don't understand the rest/

To solve these probability problems, we will use the cumulative distribution function (CDF) of the normal distribution. The CDF gives us the probability that a random variable X takes on a value less than or equal to a given value.

For the first problem, we want to find the percentage of female polychaete worms that have lengths between 7 mm and 8 mm. To do this, we need to standardize the values using the mean and standard deviation provided.

1. Let's standardize the lower and upper limits:
Lower limit (7 mm): (7 - 6.1) / 1.3 ≈ 0.6923
Upper limit (8 mm): (8 - 6.1) / 1.3 ≈ 1.4615

2. We can then use the CDF to calculate the probability:
P(0.6923 < X < 1.4615)

Now let's move on to the next problem:

6.9: Given that X is a normally distributed random variable with a mean of 60 and a standard deviation of 10:
a. P(X > 60):
We want to find the probability that X is greater than 60. Since we are given the mean and standard deviation, we can use the standard normal distribution.
P(X > 60) is equivalent to finding the area to the right of 60 on the standard normal curve. This is represented by P(Z > 0), where Z is a standard normal random variable with mean 0 and standard deviation 1.

b. P(60 < X < 72):
We want to find the probability that X is between 60 and 72. Again, we can use the standard normal distribution by standardizing these values.
P(60 < X < 72) is equivalent to finding the area between Z = (60 - 60) / 10 = 0 and Z = (72 - 60) / 10 = 1.2.

c. P(57 < X < 83):
We want to find the probability that X is between 57 and 83.
P(57 < X < 83) is equivalent to finding the area between Z = (57 - 60) / 10 = -0.3 and Z = (83 - 60) / 10 = 2.3.

d. P(65 < X < 82):
We want to find the probability that X is between 65 and 82.
P(65 < X < 82) is equivalent to finding the area between Z = (65 - 60) / 10 = 0.5 and Z = (82 - 60) / 10 = 2.2.

e. P(38 < X < 78):
We want to find the probability that X is between 38 and 78.
P(38 < X < 78) is equivalent to finding the area between Z = (38 - 60) / 10 = -2.2 and Z = (78 - 60) / 10 = 1.8.

f. P(X < 38):
We want to find the probability that X is less than 38.
P(X < 38) is equivalent to finding the area to the left of 38 on the standard normal curve. This is represented by P(Z < -2.2).