The average lifespan of a set of tires is 38,000 miles, with a standard deviation of 1500 miles. What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

This is the pvalue of X when X = 44000 subtracted by the pvalue of Z when X = 32000. So

X = 44000

has a pvalue of 1.

X = 32000

has a pvalue of 0.

1 - 0 = 100%

100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles

Loss% = 9%

Step-by-step explanation:

Here,

CP of both cows=Rs.10,000

Let CP of one cow be Rs.x

CP of another cow be Rs.(10,000-x)

Now,

CP=x ,loss%=30%

SP1 = CP(100 - L%)/100

= x(100 - 30)/100

= 70x/100

=7x/10

Again, CP =Rs.(10,000 - x), Profit%=30%

SP2 = CP(100 - P%)/100

= (10,000 - x)(100+30)/100

= 130 (10,000-x)/100

= (130000 - 13x)/10

According to the question,

SP1 = SP2

7x/10 = 130000 -13x

7x+13x = 130000

20x = 130000

x=6500

Then,

SP1 = (7× 6500)/100 =Rs4550

Total SP=2×4550 = Rs.9100

Since,

CP>SP, loss is made

loss = CP- SP=Rs.(10,000- 9100)=Rs.900

loss%= (loss/ CP)×100%

= (900/10,000)×100

= 9% ans

I hope it helps!!