Question

Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 109 and standard deviation 2. (a) What is the probability that chloride concentration equals 110? Is less than 110? Is at most 110? (Round your answers to four decimal places.) equals 110 less than 110 at most 110 (b) What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? (Round your answer to four decimal places.) Does this probability depend on the values of μ and σ? , this probability depend on the values of μ and σ.

Answer 1

Answer:

a) P(x = 110)  = 0

$P( x < 110) = 0.6915 = 69.15\%$

$P( x \leq 110) = 0.6915 = 69.15\%$

b) $P(109 > x > 111) = 1- 0.3413 = 0.6587 = 65.87\%$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 109

Standard Deviation, σ = 2

We are given that the distribution of  blood chloride concentration is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P( chloride concentration equals 110)

P(x = 110)  = 0

Because in a continuous distribution probability at one point is always zero.

P(chloride concentration is less than 110)

$P(x < 110) = P(z > \displaystyle\frac{110-109}{2}) = P(z < 0.5)$

Calculating the value from the standard normal table we have,

$P( x < 110) = 0.6915 = 69.15\%$

P(chloride concentration is at-most 110)

$P(x \leq 110) = P(z \leq \displaystyle\frac{110-109}{2}) = P(z < 0.5)$

Calculating the value from the standard normal table we have,

$P( x \leq 110) = 0.6915 = 69.15\%$

b) P(chloride concentration differs from the mean by more than 1 standard deviation)

P( 109 > x > 111) = 1 - P( 109 < x < 111)

$P(109 \leq x \leq 111) = P(\displaystyle\frac{109 - 109}{2} \leq z \leq \displaystyle\frac{111-109}{2}) = P(0 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < 0)\\= 0.3413 = 34.13\%$

$P(109 > x > 111) = 1- 0.3413 = 0.6587 = 65.87\%$

Clearly, this probability depend on the values of μ and σ.