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tresset_1 [31]
3 years ago
11

Question set 2: Many species of terrestrial tree frogs that hibernate at or near the ground surface can survive prolonged exposu

re to low winter temperatures. In freezing conditions, the frog’s body temperature, called its supercooling temperature, remains relatively higher because of an accumulation of glycerol in its body fluids. A study in Science revealed that the supercooling temperature (in Celsius scale,) of terrestrial frogs frozen at -6°_C has a relative frequency distribution with a mean of -2°_C and a standard deviation of 0.3°_C. Consider the mean supercooling temperature of a random sample of n = 42 terrestrial frogs frozen at -6°_C. (a) Find the probability that the mean of the sample exceeds -2.05°_C .

Mathematics
1 answer:
joja [24]3 years ago
7 0

Answer:

0.86

Step-by-step explanation:

The z-score we need to make the comparison is given by

\bf z=\frac{-2.05-(-2)}{\sigma/\sqrt{n}}

where

\bf \sigma= the standard deviation established in the study

n = the sample size

\bf z=\frac{-2.05-(-2)}{0.3/\sqrt{42}}=-1.0801

and for the Normal distribution N(0,1) we want to find  

<em>P(X > -1.0801) </em>

By using a spreadsheet or the table attached, we find that

P(X > -1.0801) = 0.86

(See picture attached)

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