What is x to the nearest tenth?

The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a

AlekseyPX

Answer:

The time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

Step-by-step explanation:

We are given that the time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a standard deviation of 40 seconds.

The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition.

Let X = time for this event for boys in secondary school

SO, X ~ Normal( $\mu=460,\sigma^{2} =40^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean time = 460 seconds

$\sigma$ = standard deviation = 40 seconds

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, it is given that the fitness association wants to recognize the fastest 10% of the boys with certificates of recognition, which means;

P(X > x) = 0.10 {where x is the required time which boy need to beat}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-460}{40}$ ) = 0.10

P(Z > $\frac{x-460}{40}$ ) = 0.10

So, the critical value of x in the z table which represents the top 10% of the area is given as 1.2816, that is;

$\frac{x-460}{40} =1.2816$

${x-460}{} =1.2816\times 40$

$x$ = 460 + 51.264 = 511.264 seconds

Hence, the time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

8 0

3 years ago

Someone please help I want to double check PLEASEEEEE HELLPPPP

nikitadnepr [17]

Answer:

correct

Step-by-step explanation:

6 0

3 years ago

Gianna is making up bags of fruit for her soccer team. Each bag contains 2 strawberries, 4 blueberries, and 3 raspberries. If sh

NeTakaya

Answer:

Gianna uses 24 raspberries

Step-by-step explanation:

there are 2 strawberries in each bag, if Gianna uses 16 then she must have 8 bags.

So 8 times 3 is 24 (The 3 is how many raspberries in a bag)

So Gianna uses 24 raspberries

Hope this helps!

7 0

3 years ago

An accessories store sells $8 custom socks and $11 custom hats. If 11 items were sold on Monday, and the

Vanyuwa [196]

Answer:

7 socks were sold

Step-by-step explanation:

7*8=56

11-7=4

4×11=44

56+44=100

This proves the answer of 7 to be correct

7 0

3 years ago

The sale bin in a clothing store contains an assortment of t-shirts in different sizes. There are 7 small, 8 medium, and 4 large

umka21 [38]

Answer:

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

Step-by-step explanation:

We have 7 small shirts, 8 medium shirts and 4 large shirts

Total number of shirts = 7 + 8 + 4 = 19 shirts

The probability that at least one of the first four shirts he checks is a large is given by

P(at least 1 large) = 1 - P(no large)

So first we need to find the probability that the none of the first four shirts he checks are large.

For the first check, there are 15 small and medium shirts and total 19 shirts so,

15/19

For the second check, there are 14 small and medium shirts and total 18 shirts left so,

14/18

For the third check, there are 13 small and medium shirts and total 17 shirts left so,

13/17

For the forth check, there are 12 small and medium shirts and total 16 shirts left so,

12/16

the probability of not finding the large shirt is,

P(no large) = 15/19*14/18*13/17*12/16

P(no large) = 0.352

Therefore, the probability of finding at least one large shirt is,

P(at least 1 large) = 1 - P(no large)

P(at least 1 large) = 1 - 0.352

P(at least 1 large) = 0.648

P(at least 1 large) = 64.8%

4 0

3 years ago