Question

i can determine when a data set warrants a normal distribution. i can determine the mean and standard deviation of a data set and fit it to a normal distribution. i can estimate population percentages based on mean, standard deviation, and distribution. i can estimate the areas under the normal curve using calculators, spreadsheets, and tables. the histogram gives data about a random sample of 15 babies born at a hospital and measures their length in inches. the mean is 20 and the standard deviation is 1.5. use this information to answer questions 3 - 6. baby length (in.) 3. what is the shape of the data distribution shown by the histogram? (lesson 22.2) (1/2 point) a. normal b. skewed left c. skewed right 4. the interval of babies length within 1 standard deviation of the mean is 18.5 < x < 21.5. according to the graph, how many of the babies have a length within this range of inches? (lesson 22.2) (1 point) a. 3 babies b. 7 babies c. 10 babies d. 14 babies 5. determine the percent of the data that are within 1 standard deviation of the mean. (lesson 22.2) (1 point) a. 20% b. 47% c. 67% d. 94% 6. suppose one of the babies is chosen at random. which fraction shows the probability that the baby has a length more than 1 standard deviation above the mean? (lesson 22.2) (1 point) a. 1/15 b. 1/5 c. 2/5 for questions 7 - 9, use the normal distribution model to help your answers. lifelong algebra 2b lep 5 l4l - algebra 2b (2020) page 156

Explanation:

Step1: Analyze the shape of the histogram

A normal distribution is symmetric. A skewed - left distribution has a longer tail on the left, and a skewed - right distribution has a longer tail on the right. Looking at the histogram, the data is not symmetric. The tail is on the left side (fewer data points on the left side for lower lengths like 17 and 18 compared to the right side for higher lengths like 21 and 22). So it is skewed left.

Step2: Calculate the number of babies within 1 standard deviation

The interval is \(18.5\lt x\lt21.5\). Count the number of bars (frequency) for \(x = 19\), \(x = 20\), \(x=21\). The frequency for \(x = 19\) is 4, for \(x = 20\) is 3, for \(x = 21\) is 3. So \(4 + 3+3=10\)

Step3: Calculate the percentage within 1 standard deviation

The total number of babies \(n = 15\). The number of babies within 1 standard deviation is 10. The percentage is \(\frac{10}{15}\times100\%\approx67\%\)

Step4: Calculate the probability more than 1 standard deviation above the mean

The mean \(\mu = 20\), standard deviation \(\sigma=1.5\). More than 1 standard deviation above the mean is \(x\gt20 + 1.5=21.5\). The number of babies with \(x\gt21.5\) (i.e., \(x = 22\)) is 3. The probability \(P=\frac{3}{15}=\frac{1}{5}\)

Answer:

1. B. Skewed left
2. C. 10 babies
3. C. 67%
4. B. \(\frac{1}{5}\)