Question

The attendance of Football games at a certain stadium is normally distributed with a mean attendance of 45000 and standard deviation of 3000.
i) What percentage of the time should be between 44,000 and 48,000? ii) What is the probability of exceeding 50,000?

Answer 1

Answer:

Assuming normal distribution

Standardized cut-off values

z1 = (48000-45000)/3000=1.0

z2 = (44000-45000)/3000= - 0.33

Area of normal curve between - 0.33 & 1

= 0.1293 + 0.3413 = 0.4706

Probability of attendance between 44,000 & 48,000 = 0.4706

Step-by-step explanation:

Im sure this is wrong but maybe try it?