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Answer:
a) 0.00262 is the fraction of patients expected to have a fever greater than 103.58 F.
b) 0.748 is the fraction of people expected to have temperature between 101.43 F and 102.74 F.
Step-by-step explanation:
We are given the following information:
n = 250
Population mean, μ = 102.00 F
Standard Deviation, σ = 0.5660
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(x > 103.58)
Converting into standard normal variate
[tex]P(z > \displaystyle\frac{103.58 - 102}{0.5660})\\\\= P(z > 2.791)\\= 1 - p(z \leq 2.791)\\= 0.00262[/tex]
The probability value is calculated from normal standard table.
Thus, 0.00262 is the fraction of patients expected to have a fever greater than 103.58 F.
b)P(101.43 < x < 102.74)
Converting into standard normal variate
[tex]P(\displaystyle\frac{101.43 - 102}{0.5660} < z < \displaystyle\frac{102.74 - 102}{0.5660})\\\\= P( -1.007 < z < 1.307)\\= P(z < 1.307) - P(z < -1.007)\\= 0.904 - 0.156\\=0.748[/tex]
The probability value is calculated from normal standard table.
Hence, 0.748 is the fraction of people expected to have temperature between 101.43 F and 102.74 F.
Gaussian distribution, bell shaped distribution, normal distribution, all are same. The values needed for the given distribution of temperatures are:
- The fraction of patients expected to have a fever greater than 103.58°F is 0.0026(probability) or 0.26%(percent) or 26/10000(fraction) approx
- The fraction of patients expected to have a temperature between 101.43°F and 102.74°F is 0.7487 (probability) or 74.87%(percent) or 7487/10000(fraction) approx.
How to get the z scores?
If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.
If we have
[tex]X \sim N(\mu, \sigma)[/tex]
(X is following normal distribution with mean [tex]\mu[/tex]standard deviation [tex]\sigma[/tex])
then it can be converted to standard normal distribution as
[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]
(Know the fact that in continuous distribution, probability of a single point is 0, so we can write
[tex]P(Z \leq z) = P(Z < z) )[/tex]
Also, know that if we look for Z = z in z tables, the p value we get is
[tex]P(Z \leq z) = \rm p \: value[/tex]
For the given case, let the random variable X be tracking the temperature readings(in degree Fahrenheit) of flu patients for given clinic in given condition. Then, by the specified data, we get:
[tex]X \sim N(102.00, 0.566)[/tex]
The fraction of patients can be expressed in percentage or probability.
Calculating the needed fractions:
a) Fraction of patients having temperature > 103.58°F = P(X > 103.58°F)
Using the standard normal distribution transformation, we get:
P(X > 103.58°F) = [tex]P(Z > \dfrac{103.58 - \mu}{\sigma}) = P(Z > \dfrac{103.58 - 102}{0.566}) \approx P(Z > 2.79)[/tex]
P(Z > 2.79) = 1 - P(Z ≤ 2.79)
From z tables, for z = 2.79, the p value is
Thus,
P(X > 103.58°F) = 1 - P(Z ≤ 2.79) = 1 - 0.9974= 0.0026
b) The fraction of patients expected to have a temperature between 101.43°F and 102.74°F = P(101.43°F≤ X ≤ 102.74°F)
P(101.43°F≤ X ≤ 102.74°F) = P( X ≤ 102.74°F) - P(X < 101.43°F)
Using standard normal distribution and z tables, we get:
[tex]P(X \leq 102.74) = P(Z \leq \dfrac{102.74 - 102}{0.566}) \approx P(Z < 1.31) =0.9049[/tex]
and
[tex]P(X < 101.43) = P(X \leq 101.43) = P(Z \leq \dfrac{101.43 -102}{0.566}) \approx P(Z \leq -1.01)\\\\P(X < 101.43) \approx 0.1562[/tex]
Thus,
P(101.43°F≤ X ≤ 102.74°F) = P( X ≤ 102.74°F) - P(X < 101.43°F)
= 0.9049 - 0.1562 = 0.7487
Thus,
The values needed for the given distribution of temperatures are:
- The fraction of patients expected to have a fever greater than 103.58°F is 0.0026(probability) or 0.26%(percent) or 26/10000(fraction) approx
- The fraction of patients expected to have a temperature between 101.43°F and 102.74°F is 0.7487 (probability) or 74.87%(percent) or 7487/10000(fraction) approx.
Learn more about standard normal distribution here:
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