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?
Your answer

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

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?
Your 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.

dnji

step-by-step explanation:

dskfbipo

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg