Question

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.

Answer 1

Answer:

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  =  $\frac{X-\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = population mean = 6 ounces

            $\sigma$ = 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( $\frac{X-\mu}{\sigma}$ < $\frac{5-6}{1.4}$ ) = P(Z < -0.71) = 1 - P(Z $\leq$ 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.