The probability that a selected fruity chew length is between 0.8 and 1.5 cm is 0.00578.
A fruity chew's length is normally distributed with random variable X with an average width of 1.2 cm.
The standard deviation is 0.12 cm.
Therefore, mean μ = 1.2 cm and;
Standard deviation, σ = 0.12 cm
Now, the probability is given as:
z = ( x - μ ) / σ
The probability that the length of a selected fruity chew is between 0.8 and 1.5 cm.
P( 0.8 < x < 1.5 )
For, P( X > 0.8)
z = ( x - μ ) / σ
z = ( 0.8 - 1.2 ) / 0.12
z = - 0.4 / 0.12
z = - 3.33
Using the z score table,
P( X > 0.8 ) = 0.99957
For P( x < 1.5),
z = ( x - μ ) / σ
z = ( 1.5 - 1.2 ) / 0.12
z = 2.5
Using the z score table,
P( X < 1.5) = 0.99379
Therefore, the probability is between 0.8 and 1.5 cm will be:
= 0.99379 - 0.99957 = 0.00578
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