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I am practising normal distribution exam type questions but I am stuck at this one:

The masses of individual apples sold in a food store are normally distributed. The supplier who provides the store with apples knows that 75% of the apples have a mass greater than 85 grams and that 10% of the apples have a mass greater than 120 grams.

(a) Find the value of the mean and the value of the standard deviation

So I successfully found the mean (μ=97.1) and the standard deviation (δ=17.9) by calculating the z-scores of the given probabilities.

The apples are always sold in bags containing 6 apples.

So how do I find the following probabilities for the grouped items? Do I have to use binomial distribution (6 trials) or something else?

(b) Find the probability that each apple in a randomly selected bag has a mass less than 105 grams.

(c) How many apples (to the nearest whole number) from a randomly selected bag would you expect to have a mass greater than 90 grams?

Thank you for the guidance in advance!

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  • $\begingroup$ Yes (b) and (c) involve a binomial distribution $\endgroup$ – Henry Mar 22 at 17:45
  • $\begingroup$ @Henry but how do I get the number of trials? $\endgroup$ – Fluellen Mar 22 at 18:45
  • $\begingroup$ For the binomial distribution, the number of trials is $n=6$ as you suggest. You have to calculate the $p$ parameters from "less than $105$ grams" or "greater than $90$ grams" $\endgroup$ – Henry Mar 22 at 20:52
  • $\begingroup$ @Henry ok, so the no. of trials is n=6 and I calculated the p to be p=0.67, but still I would need an x variable, that is the no. of successful outcomes. $\endgroup$ – Fluellen Mar 23 at 9:18
  • $\begingroup$ So for each of the six apples to have a mass below $105$, you need $X=6$ in the binomial, which gives a probability of about ${6 \choose 6}0.67^6 0.33^0= 0.67^6 \approx 0.09$ or $9\%$ $\endgroup$ – Henry Mar 23 at 10:07
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So the b part can be solved using a binomial distribution:

${6 \choose 6}0.67^6 0.33^0= 0.67^6 \approx 0.09$

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