The number of chips in a bag is normally distributed with a mean of 74 and a standard deviation of 4. Approximately what percent of bags contain between 62 and 86 chips

99.73% of bags contain between 62 and 86 chips .

Step-by-step explanation:

We are given that the number of chips in a bag is normally distributed with a mean of 74 and a standard deviation of 4.

Let X = percent of bags containing chips

So, X ~ N()

The standard normal z score distribution is given by;

         Z = ~ N(0,1)

So, percent of bags contain between 62 and 86 chips is given by;  

P(62 < X < 86) = P(X < 86) - P(X <= 62)                              

P(X < 86) = P( < ) = P(Z < 3) = 0.99865 {using z table}

P(X <= 62) = P( <= ) = P(Z <= -3) = 1 - P(Z < 3)= 1 - 0.99865 = 0.00135

So, P(62 < X < 86) = 0.99865 - 0.00135 = 0.9973 or 99.73%

Therefore, 99.73% of bags contain between 62 and 86 chips .

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step-by-step explanation: