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

step-by-step explanation:

let

we know that

the y-intercept is the y-coordinate of the point when the value of x is equal to zero

so

the y-coordinate of the point a is the y-intercept

the answer part a) is

the value of y-intercept is

part b) find the slope m

we know that

the value of the slope is equal to

substitute the values

therefore

the answer part b) is

the value of the slope is equal to

the general answer is

the slope is

the value of y-intercept is

probably the last option

step-by-step explanation: