Question: 1 pt 11of12?complete) ? This study finds that the carapace length of a

Question

Question: 1 pt 11of12?complete) ? This study finds that the carapace length of an adult spider is normally distributed with a mean of 18.03 mm and a standard deviation of 1.53 mm. Let x denote carapace length for the adult spider Sketch the distribution of the variable x. Choose the correct graph below ??. a. 153 38 56 315 1803 2123 b. Obtain the standardized version, z, of x. Choose the correct standardized version below OA. (z-1803) OB. ,(K-18.03) - 218 03 o. identify and sketch the distribution of z Choose the corredt graph below ??. Click to select your answerts)

Explanation / Answer

Solution:-
a. option C.

b. option B. Z = (x - 18.03)/1.53

c. option A.

d. P(15 < x < 16) = (15 - 18.03)/1.53 < Z < (16 - 18.03)/1.53
= -1.9804 < Z < -1.3268

The percentage of adult spiders that have carapace length between 15mm and 16 mm is equal to the area under the standard normal curve between -1.98 and -1.33

e. P(X > 19) = P(Z > (19 - 18.03)/1.53)
= P(Z > 0.6340)
= 0.2643

The percentage of adult spiders that have carapace length exceceding 19 mm is equal to the area under the standard normal curve that lies to the right of 0.2643

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