The weights of adult male beagles are normally distributed, with a mean of 25 pounds and a standard deviation of 4 pounds.
if 75 beagles are randomly selected, then approximately how many are expected to weigh more than 30 pounds?
a)8   
b)15
c) 30
d) 12
e) 25

A) 8

Step-by-step explanation:

Mean weight = u = 25 pounds

Standard Deviation = = 4 pounds

We have to find how many beagles out of 75 will weigh more than 30. Since the data is normally distributed, we can use z score to find this value. First we will what is the percentage(probability) of a randomly selected beagle to weigh more than x = 30 pounds, using this percentage we can then find our answer.

The formula for z score is:

Using the values, we get:

So, P(Weight > 30) is equivalent to P(z > 1.25). Using the z table, we can write:

P(z > 1.25) = P(Weight > 30) = 0.1056

This, 0.1056 or 10.56% of the beagles are expected to weigh more than 30 pounds.

So,out of 75 beagles, 10.56% of 75 are expected to weigh more than 30 pounds.

10.56% of 75 = 0.1056 x 75 = 7.92 = 8 (rounding to nearest integer)

Therefore, out of 75 beagles 8 are expected to weigh more than 30 pounds.

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!