Question

3. A candy bar producer is producing a new candy bar that has a normal distribution with a mean of 43.4 grams and a standard deviation of.15 grams. a. If every bar is labeled as 43 grams, what proportion of the bars contain less than the labeled amount? b. If only 1.5% of the candy bars exceed weight w, what is the value of w?

Answer 1

Hello Sir/ Mam

Given that:

Candy Bar's weight has a normal distribution with mean = 43.4 grams and Standard Deviation = 0.15grams

(a) THE REQUIRED PROPORTION = 0.38%

$P(X<43)=P(z<\frac{43-43.4}{0.15})=P(z<-2.67)=0.00383 =0.38\%$

(b) w = 43.73

Given that 1.5% of the candy bar exceed weight w.

Hence, the 98.5% percentile value will give us the weight w.

Now, using excel function, "=NORMSINV(0.985)", we can easily get the z-score = 2.17

Hence,

$2.17=\frac{X-43.4}{0.15}$

$0.3255=X-43.4$

$X=43.73$

I hope this solves your doubt.

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