Clarissa works in sales for a software company. Her pay each month varies, because each month she receives a base pay of $2,400.
1 answer:
Answer:
Total earning of Clarissa for the november month is $3840 .
Step-by-step explanation:
As given
Clarissa works in sales for a software company.
Her pay each month varies, because each month she receives a base pay of $2,400.00 plus 8% commission on her sales.
During the month of November, Clarissa had $18,000.00 in sales.
8% is written in the decimal form
= 0.08
Commission price = 0.08 × sales in november month .
= 0.08 × 18000
= $ 1440
Total earning in november month = Base price + Commission price
= $2400 + $ 1440
= $ 3840
Therefore the total earning of Clarissa for the november month is $3840 .
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Answer:
And we can use the z score formula given by:
And if we find the z score for 76 we got:
And if we use the complement rule and the normal standard distribution or excel we got:
And we can find this using the ti 84 with the following code:
1-normalcdf(-1000, 76, 75, 0.5 )
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Solution to the problem
Let X the random variable who represent the score of golfers. We select a sampel size of 81.
From the central limit theorem we know that the distribution for the sample mean is given by:
And we want to find this probability:
And we can use the z score formula given by:
And if we find the z score for 76 we got:
And if we use the complement rule and the normal standard distribution or excel we got:
And we can find this using the ti 84 with the following code:
1-normalcdf(-1000, 76, 75, 0.5 )