Question

Please explain steps thoroughly of how one would solve this using technology. Is there also a way to avoid using the standard normal table or is it necessary? thanks!

A particular fruit's weights are normally distributed, with a mean of 456 grams and a standard deviation of 7 grams. If you pick 21 fruits at random, then 16% of the time, their mean weight will be greater than how many grams? Give your answer to the nearest gram.

Answer 1

Solution:-

Given that,

mean = $\mu$ = 456

standard deviation = $\sigma$ = 7

n = 21

$\mu$$\bar x$ =$\mu$ = 456

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 7/ $\sqrt$ 21 = 1.5275

P(Z > z ) = 0.16

1- P(z < z) =0.16

P(z < z) = 1-0.16 = 0.84

z = 0.99

Using z-score formula,

$\bar x$ = z * $\sigma$ $\bar x$ + $\mu$ $\bar x$  = 0.99*1.5275 + 456 = 457.51

$\bar x$ = 458 gram