Question

he weights of apples are normally distributed with a mean weight of 100 grams and standard deviation of 15 grams. If an apple is selected at random, what is the probability that its weight is:

PLEASE USE THE "NORMDIST" FUNCTION IN EXCEL

a) more than 111 grams;

b) between 97 and 107 grams;

c) at most 99 grams;

d) less than 98 or more than 105 grams;

e) exactly 97.8 grams;

f) exactly 87.8 or 106.9 grams;

g) not 112 grams;

h) at least 117 grams.

Answer 1

Solution:

Given:

The weights of apples are normally distributed with a mean weight of 100 grams and standard deviation of 15 grams.

We have to use NORMDIST in Excel to find following probabilities:

Part a)
P( more than 111 grams ) = ......?

P( X> 111 ) =............?

P( X > 111) = 1 - P( X < 111 )

Thus we use following Excel command:

=1-NORMDIST( x , mean , SD , cumulative)

=1-NORMDIST(111,100,15,TRUE)

=0.2317

Thus P( more than 111 grams ) = 0.2317

Part b)

P( between 97 and 107 grams) = ......?

P( 97 < X < 107) = .......?

P( 97 < X < 107) = P( X < 107 ) - P( X < 97)

Thus

=NORMDIST( Upper x , mean , SD , Cumulative) - NORMDIST( Lower x , mean , SD , Cumulative)

=NORMDIST(107,100,15,TRUE)-NORMDIST(97,100,15,TRUE)

=0.2589

Thus P( between 97 and 107 grams) = 0.2589

Part c)

P(  at most 99 grams ) = ..............?

P( X 99 ) = .......?

Thus

=NORMDIST( x , mean , SD , cumulative)

=NORMDIST(99,100,15,TRUE)

=0.4734

Thus P(  at most 99 grams ) = 0.4734

Part d)

P( less than 98 or more than 105 grams ) = .......?

P( X < 98 or X > 105) =.........?

P( X < 98 or X > 105) = P( X < 98) + P( X > 105)

P( X < 98 or X > 105) = P( X < 98) + [ 1 - P( X < 105) ]

Thus

=NORMDIST( x , mean , SD , cumulative) - ( 1 - NORMDIST( x , mean , SD , cumulative) )

=NORMDIST(98,100,15,TRUE)-(1-NORMDIST(105,100,15,TRUE))

=0.0775

Thus P( less than 98 or more than 105 grams ) = 0.0775

Part e)

P( exactly 97.8 grams) =........?

P( X = 97.8) =......?

Thus

=NORMDIST( x , mean , SD , cumulative)

=NORMDIST(97.8,100,15,FALSE)

( note here we use FALSE for cumulative , since we have to find probability of X = 97.8)

=0.0263

Thus P( exactly 97.8 grams) =0.0263

Part f)

P( exactly 87.8 or 106.9 grams)= ........?

P( X = 87.8 or X = 106.9 ) = ..........?

P( X = 87.8 or X = 106.9 ) = P( X = 87.8) + P( X = 106.9 )

Thus

=NORMDIST( x , mean , SD , cumulative) + NORMDIST( x , mean , SD , cumulative)

=NORMDIST(87.8,100,15,FALSE)+NORMDIST(106.9,100,15,FALSE)

=0.0430

Thus P( exactly 87.8 or 106.9 grams)= 0.0430

Part g)

P(  not 112 grams) = ..............?

P(  not 112 grams ) =1 - P( X = 112)

Thus

=1-NORMDIST( x , mean , SD , cumulative)

=1-NORMDIST(112,100,15,FALSE)

=0.9807

Thus P(  not 112 grams) = 0.9807

Part h)

P( at least 117 grams ) = .......?

P( X 117 ) = 1 - P( X < 117)

Thus

=1-NORMDIST( x , mean , SD , TRUE)

=1-NORMDIST(117,100,15,TRUE)

=0.1285

Thus P( at least 117 grams ) = 0.1285