The loads carried by an elevator are found to follow a normal distribution with a mean weight of 1812 lbs, and a standard deviation of 105.3 lbs. in which interval centered about the mean does the load lie, in 95% of all cases? a) [1606, 2018] b) [1606, 1812] c) [1812, 2018] d) [1602, 2000]

No one of the mentioned

The interval is:

[ 1706.7 : 1917.3]  to find 95.5 % of all values

Step-by-step explanation:

We deal with a Normal Distribution, : μ = 1812 and σ = 105.3

We know  centered at mean value and at both sides at values σ/2 we will find 68.3 % of all values . In fact σ/2 to the left and σ/2 to the right means an interval μ + σ .

[μ + σ/2    :      μ  -  σ/2 ]      will lie 68.3 % of values

Following the same reasoning but now for a wider interval

[μ + σ   :    μ  -  σ ]  We will find 95.5 % of all values in our case that interval is:

1812 + 105,3  = 1917.3

1812 - 105,3   = 1706.7

[ 1706.7 : 1917.3] will be the interval to find 95.5 % of all values