normally distributed, with a mean of 431 grams and a standard
deviation of 29 grams.
If you pick 10 fruit at random, what is the probability that their
mean weight will be between 405 grams and 453 grams
P(405<¯x<453)=P(405<x¯<453)= Incorrect
Xbar = mean weight
P(405 < xbar < 453)
Here mean = 431
And sd=standard deviation = 29
And sample size = n = 10
(xbar - mean) /(SD /n) follows stand normal distribution N(0,1)
Hence (xbar - 431)/(29/10) follows N(0,1)
P(405 < xbar < 453)
= P{ (405 - 431) /(29/10) < N(0,1) <
(453 - 431) /(29/10) }
= P(-2.84 < N(0,1) < 2.40)
= {P( N(0,1) < 2.40)} - P(N(0,1) < - 2.84)
= {1- P(N(0,1) > 2.40)} - P(N(0,1) > 2.84)
= 1 - 0.0082 - 0.0023
= 0.9895
normally distributed, with a mean of 431 grams and a standard deviation of 29 grams. If...
A particular fruit's weights are normally distributed, with a mean of 431 grams and a standard deviation of 29 grams. If you pick 10 fruit at random, what is the probability that their mean weight will be between 405 grams and 453 grams P(405<¯x<453)=P(405<x¯<453)=
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