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?

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%

there are no figures how am i supposed to answer this question

step-by-step explanation:

i can

step-by-step explanation: