Question

The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 257.62 and a standard deviation of 62.1 (All units are 1000 cells/μ L.)
Using the empirical rule, find each approximate percentage below.
What is the approximate percentage of women with platelet counts within 1 standard deviation of the mean, or between 195.5 and 319.7 ?
What is the approximate percentage of women with platelet counts between 71.3 and 443.9 ?
a. Approximately 68 % of women in this group have platelet counts within 1 standard deviation of the mean, or between 195.5 and 319.7.
(Type an integer or a decimal. Do not round.)
b. Approximately ____ % of women in this group have platelet counts between 71.3 and 443.9.
(Type an integer or a decimal. Do not round.)

Answer 1

Answer:

(a) Approximately 68 % of women in this group have platelet counts within 1 standard deviation of the mean, or between 195.5 and 319.7.

(b) Approximately 99.7% of women in this group have platelet counts between 71.3 and 443.9.

Step-by-step explanation:

We are given that the blood platelet counts of a group of women have a bell-shaped distribution with a mean of 257.62 and a standard deviation of 62.1

Let X = the blood platelet counts of a group of women

So, X ~ Normal($\mu=257.62, \sigma^{2} =62.1^{2}$)

Now, the empirical rule states that;

• 68% of the data values lie within the 1 standard deviation of the mean.

• 95% of the data values lie within the 2 standard deviations of the mean.

• 99.7% of the data values lie within the 3 standard deviations of the mean.

(a) The approximate percentage of women with platelet counts within 1 standard deviation of the mean, or between 195.5 and 319.7 is 68% according to the empirical rule.

(b) The approximate percentage of women with platelet counts between 71.3 and 443.9 is given by;

       z-score of 443.9 =  $\frac{X-\mu}{\sigma}$

                                   =  $\frac{443.9-257.62}{62.1}$  = 3

       z-score of 71.3 =  $\frac{X-\mu}{\sigma}$

                                =  $\frac{71.3-257.62}{62.1}$  = -3

So, approximately 99.7% of women in this group have platelet counts between 71.3 and 443.9.