Question

The weight of a basketball is normally distributed with a mean of 17 oz and a standard deviation of 2 oz. Suppose 500 different basketballs are in a warehouse. About how many basketballs weigh more than 19 oz?

Answer 1

79 balls.

Explanation:

We have been given that the weight of a basketball is normally distributed with a mean of 17 oz and a standard deviation of 2 oz.

Let us find the z-score for the weight 19 oz.

`z=(x-\mu)/(\sigma)\\\\z=(19-17)/(2)\\\\z=(2)/(2)\\\\z=1`

Let us find P(z > 1) using normal distribution table.

P(z > 1) = 1 - 0.84134

= 0.15866

So the probability of a basketball having weight more than 19 oz is 0.15866. As there are 500 basketballs in the warehouse, so the total number of basketballs having a weight more than 19 oz will be :

Total number of balls having weight more than 19 oz = 500 × 0.15866

= 79.33

≈ 79 balls

Therefore, 79 balls weigh more than 19 oz

Answer 2

Answer: Hello!

In a normal distribution, the proportion between the mean and the mean plus the standard deviation is 34.1%, between the mean plus the standard deviation, and the mean plus 2 times the standard deviation is 13.6%, between the mean plus 2 times the standard deviation and the mean plus 3 times the standard deviation you have a 2,1%, and there is another 0.2% for upper values (because the total addition should be 50% if we are looking only to half the distribution).

Then, we have 500 balls, where the mean is 17 oz, and the standard deviation is 2 oz, and we want to calculate the amount of balls with more than 19 oz, this is calculate the amount of balls with more than the mean plus one time the standard deviation.

this is 13.6% +2.1% + 0.2% = 15.9%

if we want to write this number as a decimal, we need to divide it by 100.

this is 0.159 which is the proportion of balls with more than 19oz.

then 0.159*500 = 79.5 balls.

rounding up, you should expect to have 80 balls with more than 19oz.