Question

A particular fruitâs weights are normally distributed, with a mean of 598 grams and a standard deviation of 6 grams. The heaviest 8% of fruits weigh more than how many grams? Give your answer to the nearest gram.

Answer 1

Answer:

The 8% of the fruit weigh more than $x=606.43 \ g$

Step-by-step explanation:

From the question we are told that

  The mean is  $\mu = 598 \ g$

    The standard deviation is $\sigma = 6 \ g$

Generally the 8% is mathematically represented as

    $P(X > x) = 0.08$

=>   $P(X > x) = P ( \frac{X - \mu}{\sigma }>\frac{x - 598}{6} )=0.08$

$\frac{X -\mu}{\sigma }  =  Z (The  \ standardized \  value\  of  \ X )$

       $P(X > x) = P ( Z >\frac{x - 598}{6} )=0.08$

From the normal distribution table the critical value corresponding  area representing 0.08 towards the right tail of the curve is

       $z = 1.405$

So

       $\frac{x- 598}{6} = 1.405$

=> $x=606.43 \ g$