confused about normal distribution s2 question

a biologist finds a nesting colony of a previously unknown sea bird on a remote island. she is able to take measurements on 100 eggs before replacing them in their nests. she records their weights, wg, in this frequency table:

weight (w): 25<w<_27 27<w<_29 29<w<_31 31<w<_33 33<w<_35
frequency: 2 13 35 33 17

and 35<w<_37 has f = 0.

mean = 31
standard deviation = 1.98

1. Assuming the weights of the eggs for this type of bird are Normally distributed and that their mean and standard deviation are the same as those of the sample, find how many eggs you would expect to be in each of these categories.

2. Do you think the assumption that the weights of the eggs are Normally distributed is reasonable?

Reply 1

grace_

1. X = weight of egg. X~N(31, 1.98²)
Find P(25<X<27) using

z = \frac{x-\mu}{\sigma}

Multiply that by 100 to find the number of eggs you would expect between 25 and 27 grams.
Repeat for each of the other categories, and see if your results agree with those in the table. :smile:

Reply 2

danielamir2

i will only do one as an example here for the group 25<w<27
mean= 31 standard deviation=1.98
(25-31)/1.98= 3.03, (27-31)/1.98=2.02 i've ignored the negative sigh
ϕ(3.03)=0.9988 ϕ(2.02)=0.9783
ϕ(3.03) - ϕ(2.02)
0.9988 - 0.9783 =0.0205 x 100 = 2.1

(edited 8 years ago)

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