The weights of the fish in a certain lake are normally distributed with a mean of 19 lbs and a standard deviation of 6. If 25 fish are randomly selected, what is the probability that the mean weight will be greater than 17.2 lbs

0.93319

Step-by-step explanation:

We solve the question using the z score formula

z = (x-μ)/σ/√n where

x is the raw score = 17.2 lbs

μ is the population mean = 19 lbs

σ is the population standard deviation = 6

n is the random number of samples = 25 fishes

Greater than sign = >

For x > 17.2 lbs

z = 17.2 - 19/6/√25

z = 17.2 - 19/ 6/5

z = 17.2 - 19/1.2

z = -1.5

Probability value from Z-Table:

P(x<17.2) = 0.066807

P(x>17.2) = 1 - P(x<17.2)

P(x>17.2) = 1 - 0.066807

P(x>17.2) = 0.93319

Therefore, that the probability that the mean weight will be greater than 17.2 lbs is 0.93319

from the figure we get,

the x axis shows   the months and,

the y-axis shows the number of satisfied customer.

each line or each cell in upward direction (y-axis) shows 500.

the appropriate scale for the y - axis is: -

a = 2000 customers

b = 4000 customers

c = 6000 customers

d = 8000 customers

e = 10000 customers

f = 12000 customers

g = 14000 customers

two data sets with the same median will not have the same varaibility

step-by-step explanation: