Question

A particular fruit's weights are normally distributed, with a mean of 539 grams and a standard deviation of 19 grams.

If you pick 20 fruit at random, what is the probability that their mean weight will be between 533 grams and 546 grams

Answer 1

Solution:

Let X be a random variable which represents the fruit's weights.

Given that, X ~ N(539, 19²)

i.e.  μ = 539 grams and σ = 19 grams

We have to obtain P(533 < x̄ < 546). (where, x̄ is sample mean)

We know that, if X ~N(μ, σ^{​​​​​​2}) then, x̄ ~ N(μ, σ^{​​​​​​2}/n)

i.e. sampling distribution of sample mean follows normal distribution with mean μ and variance σ^{​​​​​​2}/n.

(Where, x̄ is sample mean and n is sample size.)

And if x̄ ~ N(μ, σ^{​​​​​​2}/n) then

We have, μ = 539 grams , σ = 19 grams and n = 20

Using "pnorm" function of R we get,

P(Z < 1.6476) = 0.9503 and P(Z < -1.4123) = 0.0789

The probability that mean weight of randomly selected 20 fruits will be between 533 grams and 546 grams is 0.8714.

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