Answer:
P(X>3.0) = 0.8043 or 80.43%
P(X<4.0) = 0.7673 or 76.73%
P(3.6<X<4.2) = 0.3149 or 31.49%
Step-by-step explanation:
Mean weight (µ) = 3.54 pounds
Standard deviation (σ) = 0.63 pounds
The z-score for any given weight, X, is defined as:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
a. P(X>3.0)
For X=3.0, the z-score is:
[tex]z=\frac{3.0-3.54}{0.63}\\z=-0.857[/tex]
A z-score of -0.857 falls in the 19.57-th percentile of a normal distribution, therefore, the probability that a randomly selected bunch of bananas will weigh more than 3.0 pounds is:
[tex]P(X>3.0) = 1-0.1957 = 0.8043[/tex]
b. P(X<4.0)
For X=4.0, the z-score is:
[tex]z=\frac{4.0-3.54}{0.63}\\z=0.730[/tex]
A z-score of 0.730 falls in the 76.73-th percentile of a normal distribution, therefore, the probability that a randomly selected bunch of bananas will weigh less than 4.0 pounds is:
[tex]P(X<4.0) = 0.7673[/tex]
c. P(3.6<X<4.2)
For X=3.6, the z-score is:
[tex]z=\frac{3.6-3.54}{0.63}\\z=0.095[/tex]
A z-score of 0.095 falls in the 53.78-th percentile of a normal distribution.
For X=4.2, the z-score is:
[tex]z=\frac{4.2-3.54}{0.63}\\z=1.048[/tex]
A z-score of 1.048 falls in the 85.27-th percentile of a normal distribution, therefore,the probability that a randomly selected bunch of bananas will weigh between 3.6 and 4.2 pounds is:
[tex]P(3.6<X<4.2)=0.8527-0.5378 = 0.3149[/tex]
