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brydenskl814
18.02.2020 •
Mathematics
The average weight of a particular box of crackers is 28.0 ounces with a standard deviation of 0.9 ounce. The weights of the boxes are normally distributed. a. What percent of the boxes weigh more than 26.2 ounces?
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Ответ:
a. 97.72%
Step-by-step explanation:
The weights of boxes follows normal distribution with mean=28 ounce and standard deviation=0.9 ounces.
a. We have to calculated the percentage of the boxes that weighs more than 26.2 ounces.
Let X be the weight of boxes. We have to find P(X>26.2).
The given mean and Standard deviations are μ=28 and σ=0.9.
P(X>26.2)= P((X-μ/σ )> (26.2-28)/0.9)
P(X>26.2)= P(z> (-1.8/0.9))
P(X>26.2)= P(z>-2)
P(X>26.2)= P(0<z<∞)+P(-2<z<0)
P(-2<z<0) is computed by looking 2.00 in table of areas under the unit normal curve.
P(X>26.2)=0.5+0.4772
P(X>26.2)= 0.9772
Thus, the percent of the boxes weigh more than 26.2 ounces is 97.72%
Ответ:
B C E F!
Step-by-step explanation: