Answer:
a) The 17th percentile would be 22 days
b) 97% of the values will fall within 21 and 25 days
Step-by-step explanation:
To find the first part of this question you need to know the basic normal distribution curve. The 17th percentile falls roughly at -1 standard deviations from the mean. Given that the mean is at 23 days, we know the 17th percentile would be at 22 days.
In order to solve part b, know that the normal distribution curve shows 97% of the values within - 2 and + 2 standard deviations away from the mean. This gives us values of 21 and 25 days.
The 17th percentile is "22 days" and the incubation times is "-21 to 25 days".
According to the question,
Mean,
Standard deviation,
(a)
Let,
The incubation time will be "a".
→ [tex]P(x<a) = 0.17[/tex]
→ [tex]P(\frac{x-23}{1} < \frac{a-23}{2} )= 0.17[/tex]
[tex]P(z < \frac{a-23}{1} ) = 0.17[/tex]
[tex]\Phi (a-23) = 0.17[/tex]
[tex]a -23 = \Phi (0.17)[/tex]
[tex]a = 22.046[/tex]
or,
[tex]= 22[/tex]
(b)
Let,
[tex]C_1[/tex] and [tex]C_2[/tex] be the incubation time for 97%.
→ [tex]P(-C_1<X< C_2) = 0.97[/tex]
→ [tex]P(\frac{-C_1-C_3}{1} < \frac{X-23}{1} < \frac{C_2-C_3}{1} )= 0.97[/tex]
[tex]P(-C_1-23< Z < C_2-23) = P(-1.98< Z< 1.98)[/tex]
By comparing both, we get
→ [tex]-C_1 -23 = -1.98[/tex]
[tex]-C_1 = -1.98+23[/tex]
[tex]C_1 = -21.02[/tex] or [tex]-21[/tex]
and,
→ [tex]C_2 -23 = 1.98[/tex]
[tex]C_2 =1.98+23[/tex]
[tex]C_2 = 24.98[/tex] or [tex]25[/tex]
Thus the above answers are correct.
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