Question

How did I get this wrong I need help with this problem ( please show work) P.s I don’t know what’s wrong with the quality

Answer 1

a. 0.01%

b. 99.22°F

Step-by-step explanation:

We know that the temperature follows a normal distribution with a mean of 98.18 °F and a standard deviation of 0.63 °F.

Part a.

If 100.6 °F is the lowest temperature that they consider fewer, we need to calculate the following probability

P( T > 100.6)

To calculate this probability, we first need to standardize 100.6 as follows

`\begin{gathered} z=\frac{temp\text{ - mean}}{\text{ standard deviation}} \\ \\ z=(100.6-98.18)/(0.63)=3.84 \end{gathered}`

Therefore, the probability is equivalent to

P(T > 100.6) = P(Z > 3.84)

Now, we need to use the table for positive z scores, but it gives the probability P (Z < z), so we can calculate P(Z > 3.84) as follows

P(Z > 3.84) = 1 - P(Z < 3.84)

P(Z > 3.84) = 1 - 0.9999

P(Z > 3.84) = 0.0001

Therefore, the percentage is 0.0001 x 100% = 0.01%. So 100.6 °F is appropiate because the probability that a healtly person is considered to have fewer is 0.01%

Part b.

If we want a temperature such that 5.0% exceed it, we need to find a value of z that satisfies

P(Z > z) = 0.05

P(Z < z) = 1 - 0.05

P(Z < z) = 0.95

Using the table, we get that z = 1.645. Then, we can find the temperature as follows

`\begin{gathered} \text{ temp = z}\cdot\text{ \lparen standard deviation\rparen + mean} \\ \text{ temp = 1.645\lparen0.63\rparen+98.18} \\ \text{ temp = 1.036+98.18} \\ \text{ temp = 99.22 }\degree F \end{gathered}`

Therefore, the answer for part b is 99.22°F