Suppose that the weight of sweet cherries is normally distributed with mean μ=6 ounces and standard deviation σ=1.4 ounces. What proportion of sweet cherries weigh less than 5 ounces? Round your answer to four decimal places.

The proportion of sweet cherries that weigh less than 5 ounces is 23.88%.

Step-by-step explanation:

We are given that the weight of sweet cherries is normally distributed with mean μ = 6 ounces and standard deviation σ = 1.4 ounces.

Let X = the weight of sweet cherries

SO, X ~ Normal(μ = 6 ounces , σ = 1.4 ounces)

The z-score probability distribution for the normal distribution is given by;

                                     Z  =    ~ N(0,1)

where, = population mean = 6 ounces

            = standard deviation = 1.4 ounces

Now, the proportion of sweet cherries that weigh less than 5 ounces is given by = P(X < 5 ounces)

         P(X < 5 ounces) = P( < ) = P(Z < -0.71) = 1 - P(Z 0.71)

                                                              = 1 - 0.7612 = 0.2388

The above probability is calculated by looking at the value of x = 0.71 from the z table which has an area of 0.7612.

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.

7 ft 6 in means 7 1/2, because 6 in is 1/2 of a foot. 7 ft 8 in means 7 2/3 because 8 in is 2/3 of a foot. 23/3 = 7 2/3. 15/2 = 7 1/2.

23/3   *   15/2   = 57 1/2

step-by-step explanation: