Answer:
a) P(X<50)=0.9827
b) P(X>47)=0.4321
c) P(-1.5<z<1.5)=0.8664
Step-by-step explanation:
We will calculate the probability based on a random sample of one moped out of the population, normally distributed with mean 46.7 and standard deviation 1.75.
a) This means we have to calculate P(x<50).
We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:
![z=\dfrac{X-\mu}{\sigma}=\dfrac{50-46.7}{1.75}=\dfrac{3.7}{1.75}=2.114\\\\\\P(X](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3D%5Cdfrac%7B50-46.7%7D%7B1.75%7D%3D%5Cdfrac%7B3.7%7D%7B1.75%7D%3D2.114%5C%5C%5C%5C%5C%5CP%28X%3C50%29%3DP%28z%3C2.114%29%3D0.9827)
b) We have to calculatee P(x>47).
We will calculate the z-score and then calculate the probability accordign to the standard normal distribution:
![z=\dfrac{X-\mu}{\sigma}=\dfrac{47-46.7}{1.75}=\dfrac{0.3}{1.75}=0.171\\\\\\P(X>47)=P(z>0.171)=0.4321](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3D%5Cdfrac%7B47-46.7%7D%7B1.75%7D%3D%5Cdfrac%7B0.3%7D%7B1.75%7D%3D0.171%5C%5C%5C%5C%5C%5CP%28X%3E47%29%3DP%28z%3E0.171%29%3D0.4321)
c) If the value differs 1.5 standard deviations from the mean value, we have a z-score of z=1.5
![z=\dfrac{(\mu+1.5\sigma)-\mu}{\sigma}=\dfrac{1.5\sigma}{\sigma}=1.5](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B%28%5Cmu%2B1.5%5Csigma%29-%5Cmu%7D%7B%5Csigma%7D%3D%5Cdfrac%7B1.5%5Csigma%7D%7B%5Csigma%7D%3D1.5)
So the probability that maximum speed differs from the mean value by at most 1.5 standard deviations is P(-1.5<z<1.5):
![P(-1.5](https://tex.z-dn.net/?f=P%28-1.5%3Cz%3C1.5%29%3DP%28z%3C1.5%29-P%28z%3C-1.5%29%3D0.9332-0.0668%3D0.8664)