PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!Use technology or a z-score table to answer the question.The lengths of green beans for sale at a supermarket are normally distributed with a mean of 11.2 centimeters and a standard deviation of 2.1 centimeters. Consider a bag of 150 green beans.How many green beans will be 13 centimeters or shorter?

Accepted Solution

Answer:Choice C: approximately 121 green beans will be 13 centimeters or shorter.Step-by-step explanation:What's the probability that a green bean from this sale is shorter than 13 centimeters?Let the length of a green bean be [tex]X[/tex] centimeters.[tex]X[/tex] follows a normal distribution withmean [tex]\mu = 11.2[/tex] and standard deviation [tex]\sigma = 2.1[/tex].In other words, [tex]X\sim \text{N}(11.2, 2.1^{2})[/tex], and the probability in question is [tex]X \le 13[/tex].Z-score table approach:Find the z-score of this measurement:[tex]\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143[/tex]. Closest to 0.86. Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that [tex]Z[/tex] will be less than or equal to the z-score in question. (In case the question is asking for the probability that [tex]Z[/tex] is greater than the z-score, subtract the value from table from 1.)[tex]P(X\le 13) = P(Z \le 0.857143) \approx 0.8051[/tex]."Technology" Approach Depending on the manufacturer, the steps generally include:Locate the cumulative probability function (cdf) for normal distributions.Enter the lower and upper bound. The lower bound shall be a very negative number such as [tex]-10^{9}[/tex]. For the upper bound, enter [tex]13[/tex]Enter the mean and standard deviation (or variance if required).Evaluate.For example, on a Texas Instruments TI-84, evaluating [tex]\text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 )[/tex] gives [tex]0.804317[/tex].As a result, [tex]P(X\le 13) = 0.804317[/tex].Number of green beans that are shorter than 13 centimeters:Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution. Number of trials [tex]n[/tex]: 150.Probability of success [tex]p[/tex]: 0.804317.Let [tex]Y[/tex] be the number of green beans out of this 150 that are shorter than 13 centimeters. [tex]Y\sim\text{B}(150,0.804317)[/tex].The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words, [tex]E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121[/tex]The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.