Respuesta :
Answer:
(a) Approximately 95% of women with platelet counts within 2 standard deviations of the mean.
(b) Approximately 99.7% of women have platelet counts between 65.2 and 431.8.
Step-by-step explanation:
The complete question is: The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 248.5 and a standard deviation of 61.1. (All units are 1000 cells/mul.) using the empirical rule, find each approximate percentage below.
a. What is the approximate percentage of women with platelet counts within 2 standard deviations of the mean, or between 126.3 and 370.7?
b. What is the approximate percentage of women with platelet counts between 65.2 and 431.8?
We are given that the blood platelet counts of a group of women have a bell-shaped distribution with a mean of 248.5 and a standard deviation of 61.1.
Let X = the blood platelet counts of a group of women
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 248.5
[tex]\sigma[/tex] = standard deviation = 61.1
Now, the empirical rule states that;
- 68% of the data values lie within 1 standard deviation away from the mean.
- 95% of the data values lie within 2 standard deviations away from the mean.
- 99.7% of the data values lie within 3 standard deviations away from the mean.
(a) The approximate percentage of women with platelet counts within 2 standard deviations of the mean, or between 126.3 and 370.7 is given by;
As we know that;
P([tex]\mu-2\sigma[/tex] < X < [tex]\mu+2\sigma[/tex]) = 0.95
P(248.5 - 2(61.1) < X < 248.5 + 2(61.1)) = 0.95
P(126.3 < X < 370.7) = 0.95
Hence, approximately 95% of women with platelet counts within 2 standard deviations of the mean.
(b) The approximate percentage of women with platelet counts between 65.2 and 431.8 is given by;
Firstly, we will calculate the z-scores for both the counts;
z-score for 65.2 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{65.2-248.5}{61.1}[/tex] = -3
z-score for 431.8 = [tex]\frac{X-\mu}{\sigma}[/tex]
= [tex]\frac{431.8-248.5}{61.1}[/tex] = 3
This means that approximately 99.7% of women have platelet counts between 65.2 and 431.8.
