Answer:
Step-by-step explanation:
Z = (x-µ)/(s/√n)=(98.5-98.2)/(0.62/√19) = 2.11
left tail p-value = 0.9826
so 98.26% are lower.
The probability that their mean body temperature will be less than 98.50 degree Fahrenheit is 98.26%.
A Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values.
Probability is the measure of the likelihood that an event will occur in a random experiment.
According to the given question.
Sample size, n = 19
mean, [tex]\mu = 98.20[/tex]
Observed value, x = 98.50
Standard deviation, [tex]\sigma = 0.62[/tex]
Since, we know that z-score is given by
[tex]z = \frac{x-\mu}{SE} = \frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
[tex]\implies z = \frac{98.50-98.20}{\frac{0.62}{\sqrt{19} } } =2.11[/tex]
Therefore, the required probability
P(z<2.11) = 0.9826
Hence, the probability that their mean body temperature will be less than 98.50 degree Fahrenheit is 98.26%.
Find out more information about z-score and probability here:
https://brainly.com/question/25638875
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