Can someone answer these pls
2 answers:
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Answer:
68% of buyers paid between $147,700 and $152,300.
Step-by-step explanation:
We are given that prices of a certain model of a new home are normally distributed with a mean of $150,000.
Use the 68-95-99.7 rule to find the percentage of buyers who paid between $147,700 and $152,300 if the standard deviation is $2300.
<u><em>Let X = prices of a certain model of a new home</em></u>
SO, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean price = $150,000
= standard deviation = $2,300
<u>Now, according to 68-95-99.7 rule;</u>
Around 68% of the values in a normal distribution lies between and
.
Around 95% of the values occur between and
.
Around 99.7% of the values occur between and
.
So, firstly we will find the z scores for both the values given;
Z = =
= -1
Z = =
= 1
This indicates that we are in the category of between and
.
SO, this represents that percentage of buyers who paid between $147,700 and $152,300 is 68%.