The amount of time Mary spends on the phone each month can be thought of as a random variable that is approximately normally distributed. Mary spends an average of 40 hours on her phone with a standard deviation of 4 hours. Find the probability that Mary spends between 30 and 40 hours on the phone. ​

0.4938

Step-by-step explanation:

Given that:

Mean, m = 40

Standard deviation, s = 4

Probability of spending between 30and 40 hours on phone :

Zscore = (x - mean) / standard deviation

P(Z = 40) - P(Z= 30)

((40 - 40) / 4) - ((30-40) / 4)

(0/ 4) - (-10/ 4)

P(Z < 0) - P(Z <-2.5)

P(Z^ 0).=0.5 ; P(Z < - 2.5) = 0.0062097

0.5 - 0.0062097

P = 0.4938

u rich

step-by-step explanation:

i dont know sorry

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