Question

learning goal
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.
lesson 22.2 checkpoint
once you have completed the above problems and checked your solutions, complete the lesson checkpoint
below.
complete the lesson reflection above by circling your current understanding of the learning goal.
the histogram shows a random sample of resting
heart rates (in beats per minute) for 25 adults. the
standard deviation is 8.2 and the mean is ( overline{x}=72 ).

1. what is the shape of the data distribution shown

by the histogram?
a. normal
b. skewed left
c. skewed right

1. the interval of resting heart rates within 1 standard deviation of the mean is ( 63.8<x<80.2 ). determine the

percent of the data that are within 1 standard deviation of the mean.
a. ( 88 % )
b. ( 76 % )
c. ( 48 % )
d. ( 28 % )

1. suppose one of the resting heart rates is chosen at random. which fraction shows the probability the adult

has a resting heart rate more than 1 standard deviation below the mean?
a. ( \frac{1}{25} )
b. ( \frac{3}{25} )
c. ( \frac{1}{5} )

Explanation:

1.

A normal distribution is symmetric. Skewed - left has a longer tail on the left, and skewed - right has a longer tail on the right. Looking at the histogram, it is symmetric.

Count the number of data points in the interval \(63.8\lt x\lt80.2\). The total number of data points \(n = 25\). The number of data points in the interval: \(1 + 2+4 + 7+5=19\). The percentage is \(\frac{19}{25}\times100\%=76\%\)

The number of data points more than 1 standard deviation below the mean (i.e., \(x\lt63.8\)) is \(1 + 2=3\). The total number of data points \(n = 25\). The probability is \(\frac{3}{25}\)

Answer:

A. Normal

2.