The blood platelet counts of a group of women have a bell-shaped distribution with a mean of 255.4 and a standard deviation of 61.3. (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 1 standard deviation of the mean, or between 181.5 and 320.1?
b. What is the approximate percentage of women with platelet counts between 112.2 and 389.4?

Step-by-step explanation:

A bell shaped distribution means that it is a normal distribution.

Let x be a random variable representing the blood platelet count of the group of women. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 255.4

σ = 61.3

a) the probability that a woman has platelet counts between 181.5 and 320.1 is expressed as

P(181.5 ≤ x ≤ 320.1)

For x = 181.5,

z = (181.5 - 255.4)/61.3 = - 1.21

Looking at the normal distribution table, the probability corresponding to the z score is 0.11

For x = 320.1

z = (320.1 - 255.4)/61.3 = 1.06

Looking at the normal distribution table, the probability corresponding to the z score is 0.86

Therefore,

P(181.5 ≤ x ≤ 320.1) = 0.86 - 0.11 = 0.75

The percentage of percentage of women with platelet counts within 1 standard deviation of the mean, or between 181.5 and 320.1 is

0.75 × 100 = 75%

b) the probability that a woman has platelet counts between 112.2 and 389.4 is expressed as

P(112.2 ≤ x ≤ 389.4)

For x = 112.2,

z = (112.2 - 255.4)/61.3 = - 2.34

Looking at the normal distribution table, the probability corresponding to the z score is 0.0096

For x = 389.4

z = (389.4 - 255.4)/61.3 = 2.19

Looking at the normal distribution table, the probability corresponding to the z score is 0.9857

Therefore,

P(112.2 ≤ x ≤ 389.4) = 0.9857 - 0.0096 = 0.9761

The percentage of percentage of women with platelet counts within 1 standard deviation of the mean, or between 112.2 and 389.4 is

0.9761 × 100 = 96.71%

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