Given:
The weights of dogs in a dog show is normally distributed.
Mean = 58 pounds
Standard deviation = 15
To find:
The probability that the weight of the dogs are between 35.5 and 65.5 pounds.
Solution:
We know that,
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
Where, [tex]\mu[/tex] is mean and [tex]\sigma [/tex] is the standard deviation.
We need to find the value of [tex]P(35.5<x<65.5)[/tex].
[tex]P(35.5<x<65.5)=P\left(\dfrac{35.5-58}{15}<\dfrac{x-58}{15}<\dfrac{65.5-58}{15}\right)[/tex]
[tex]P(35.5<x<65.5)=P\left(-1.5<z<0.5\right)[/tex]
[tex]P(35.5<x<65.5)=P\left(z<0.5\right)-P\left(z<-1.5\right)[/tex]
Level of significance is 95%.
[tex]P(35.5<x<65.5)=0.69146-0.066807[/tex]
[tex]P(35.5<x<65.5)=0.624653[/tex]
[tex]P(35.5<x<65.5)\approx 0.6247[/tex]
Therefore, the probability that the weight of the dogs are between 35.5 and 65.5 pounds is about 0.6247.
