Question

PLEASE HELP!!! Blue whales eat an average of 5,000 pounds of fish daily, with a standard deviation of 850 pounds. Approximately what percentage of blue whales eat more than 5,850 pounds of fish?

Answer 1

Answer: 15.87%

Explanation:

Given : Blue whales eat an average of 5,000 pounds of fish daily, with a standard deviation of 850 pounds.

i.e.
`\mu=5000\ ; \ \sigma=850`

We assume that this a normal distribution.

Let x be the random variable that shows the amount of fish eaten by whales.

z-score :
`z=(x-\mu)/(\sigma)`

For x= 5,850 pounds

`z=(5850-5000)/(850)=1`

By using standard normal distribution table , we have

The probability that the blue whales eat more than 5,850 pounds of fish :-

`P(X>5850)=P(z>1)=1-P(\leq1)\\\\=1-0.8413447=0.1586553\approx15.87\%`

Hence, the percentage of blue whales eat more than 5,850 pounds of fish = 15.87%

Answer 2

15.87% of blue whales eat more than 5,850 pounds of fish

Explanation:

Mean = 5000

Standard Deviation = 850

We need to find P(x>5850) = ?

We can find using z-score

z = x - mean/ standard deviation

z = 5850 - 5000/850

z = 850/850

z = 1

Now, P(x>5850) = P(z>1)

Finding value of P(z>1) by looking at the z-score table

P(z>1) = 0.8413

The normal distribution gives area to the left , so subtracting it from 1 to gain the answer

1-0.8413 = 0.1587

Now, to find percentage multiply it with 100

0.1587 * 100 = 15.87 %

15.87% of blue whales eat more than 5,850 pounds of fish