Answer:
Question
5) 0.81859
6) P(x<6.6) = 0.15866
7) 15.866 frogs would be longer than 8.2cm
Step-by-step explanation:
5. A biologist is measuring the lengths of frogs in a certain location. The lengths
of the frogs have a mean of 7.4 centimeters and a standard deviation of 0.8
centimeters. What is the probability that a randomly selected frog will measure
between 5.8 cm and 8.2 cm?
Using z score formula
z = (x-μ)/σ, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
z1 = (x-μ)/σ
x is the raw score = 5.8cm
μ is the population mean = 7.4cm
σ is the population standard deviation. = 0.8cm
= 5.8 - 7.4/0.8
= -2
Using a z score table to find the probability of -2
P(z = -2) = 0.02275
z2 = (x-μ)/σ,
x is the raw score = 8.2cm
μ is the population mean = 7.4cm
σ is the population standard deviation. = 0.8cm
z = 8.2 - 7.4/0.8
z = 1
Using the Z-Table to find the probability
P(z = 1) = 0.84134
Hence, the probability that a randomly selected frog will measure between 5.8 cm and 8.2 cm is calculated as:
0.84134 - 0.02275 = 0.81859
6. Using the data from question #5, what is the probability that a randomly
selected frog will measure less than 6.6 cm?
Your answer
Using z score formula
z = (x-μ)/σ, where
x is the raw score = 6.6cm
μ is the population mean = 7.4cm
σ is the population standard deviation. = 0.8cm
z1 = (x-μ)/σ
= 6.6 - 7.4/0.8
= -1
Using the Z-Table to find the probability a randomly selected frog will measure less than 6.6 cm
P(x<6.6) = 0.15866
7. Using the data from #5, if 100 frogs are measured, how many of them would you predict to measure longer than 8.2 cm?
z = (x-μ)/σ,
x is the raw score = 8.2cm
μ is the population mean = 7.4cm
σ is the population standard deviation. = 0.8cm
z = 8.2 - 7.4/0.8
z = 1
Using the Z-Table to find the probability
P(x<8.2) = 0.84134
To find the probability of frogs that would be longer than 8.2cm
P(x>8.2) = 1 - P(x<8.2) = 0.15866
To calculate the number would be predicted to measure longer than 8.2 cm is :
0.15866 × 100 = 15.866% is longer than 8.2cm
Therefore,
15.866% × 100 frogs = 15.866 frogs.
