Bananas are sold in bunches at a grocery store and typically consist of 4-8 bananas per bunch. suppose the weight of these bunches follows a normal distribution with a mean of 3.54 pounds and a standard deviation of 0.63 pounds. what is the probability that a randomly selected bunch of bananas will weigh more than 3.0 pounds? what is the probability that a randomly selected bunch of bananas will weigh less than 4.0 pounds? what is the probability that a randomly selected bunch of bananas will weigh between 3.6 and 4.2 pounds?

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:

a. P(X>3.0)

For X=3.0, the z-score is:

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:

b. P(X<4.0)

For X=4.0, the z-score is:

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:

c. P(3.6<X<4.2)

For X=3.6, the z-score is:

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:

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:

it is b)

hope this : )

step-by-step explanation:

it is x equals 3   plese mark me really need it

step-by-step explanation: