Answer: 0.1435
Step-by-step explanation:
Given : Mean = 479 pounds
Standard deviation = 40 pounds.
Let X denote the weights of cows.
[tex]X\sim N(\mu=479,\sigma=40)[/tex]
The cow transport truck holds 15 cows and can hold a maximum weight of 7350.
i.e. mean weight of cow in this case =[tex]\overline{x}=\dfrac{7350}{15}=490\text{ pounds}[/tex]
If 15 cows are randomly selected from the very large herd to go on the truck, what is the probability their total weight will be over the maximum allowed of 7350 will be:-
[tex]P(\overline{x}>490)=P(\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{490-479}{\dfrac{40}{\sqrt{15}}})\\\\=P(z>1.065)\ \ [\because\ z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-P(z\leq1.065)\\\\=1- 0.8565=0.1435\ \ [\text{By z-table}][/tex]
Hence, If 15 cows are randomly selected from the very large herd to go on the truck, the probability their total weight will be over the maximum allowed of 7350 = 0.1435
So, the probability their mean weight will be over 479 is [tex]0.53836.[/tex]
A Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. Z-score is measured in terms of standard deviations from the mean.
It is given that,
[tex]\mu=479\\\sigma=40\\n=15\\X=maximum\ weight=7350[/tex]
Then,
[tex]\bar{x}=\frac{\sum x}{n}\\ =\frac{7350}{15}\\ =490[/tex]
Now, calculating Z-score:
[tex]Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n} } } \\Z=\frac{490-479}{\frac{40}{\sqrt{15} } }\\ Z=1.07[/tex]
Using z table =1.07
[tex]P(Z < 1.07)=0.46164\\P(Z > 1.07)=1-P(Z < 1.07)\\=1-0.46164\\=0.53836[/tex]
Learn more about the topic Z-score:
https://brainly.com/question/5512053
